engineering taking a first course in ordinary differential equations. The course ...... Frank Ayres, Jr., Differential equations, Schaum's outline series, McGraw-Hill.
Ordinary Differential Equations ———————————————————————
Arbab Ibrahim Arbab (Ph.D.)
Acknowledgments This book is dedicated to many people who have been good to me. It is also dedicated to the memory of my father, who spent his life in teaching people. Last but not least, I dedicate this book to the memory of Prof. A. Salam, who remains my ideal of what a genuine scientist and humanist should be. I wish to thank Professor Al Tahir Al Agib, Gamal Al Abwabi and Badawi Mohamed for their useful comments in reading the manuscript. I would also like to thank Ms Sadia El Sir and Mr. Adil Adam for their assistance in preparing the glossary. My thanks are also due to my students and colleagues who encourage me for writing this book. I will be very grateful to the readers of this book for their anticipated feedback that will be considered in future editions.
1
Preface This book has been prepared to serve as a textbook for students in science and engineering taking a first course in ordinary differential equations. The course is intended to serve as an introduction to the theory of ordinary differential equations. The solution of differential equations with variable coefficients tackled via the method of series solution is given in chapter 5. The famous differential equations like, Bessel, Hermite, Legendre, Laguerre, Hypergeometric equations, are either solved or given as an exercise. Some physical applications to differential equations are given in chapter 6. These examples elucidates the importance of the differential equations in physics and related sciences. An appendix containing standard integrals is put at the end of the book. To help the students who had the subject in Arabic, we present a glossary for all words appearing in the book.
April 2001, Khartoum
Arbab, A.I
2
ﺑﺴﻢ اﻟﻠّﻪ اﻟﺮﺣﻤﻦ اﻟﺮﺣﯿﻢ
ﻤﻘﺩﻤـﺔ ، ﺍﻟﻤﺅﻟﻑ ﺃﺭﺒﺎﺏ ﺇﺒﺭﺍﻫﻴﻡ ﺃﺭﺒﺎﺏ
Contents 1 Differential Equations 1.1
1.2
1
The origin of differential equations . . . . . . . . . . . . . . . . .
1
1.1.1
The order of a D.E. . . . . . . . . . . . . . . . . . . . . .
1
1.1.2
The degree of a D.E . . . . . . . . . . . . . . . . . . . . .
2
1.1.3 The solution of differential equations . . . . . . . . . . . . The general and particular solutions of the D.Es. . . . . . . . . . .
5 6
1.2.1
First order differential equations . . . . . . . . . . . . . . . 11
1.2.2
Homogeneous Equations . . . . . . . . . . . . . . . . . . . 15
1.2.3
Equations reducible to homogeneous equation . . . . . . . 19
2 Exact Differential Equations 2.1
21
2.0.4 Exact differential equation . . . . . . . . . . . . . . . . . . 21 Solution of an Exact D.E. . . . . . . . . . . . . . . . . . . . 24
3 First Order Linear Differential Equations
29
3.0.1
Linear Differential Equations . . . . . . . . . . . . . . . . . 29
3.0.2
Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . 30
3.0.3 3.0.4
First order D.Es. with higher degrees . . . . . . . . . . . . 34 Clairaut’s Equation . . . . . . . . . . . . . . . . . . . . . 37
3.0.5
Ricatti’s Equation . . . . . . . . . . . . . . . . . . . . . . 39
3.0.6
Lagrange’s Equation . . . . . . . . . . . . . . . . . . . . . 40
i
4 Higher Order Linear Differential Equations 43 4.1 Homogeneous D.Es. with constant coefficients . . . . . . . . . . . 45 4.1.1 4.2
The complementary solution of a D.E. . . . . . . . . . . . 47
Non homogeneous linear D.Es. . . . . . . . . . . . . . . . . . . . 50 4.2.1 4.2.2
Operator Method . . . . . . . . . . . . . . . . . . . . . . 50 The properties of L(D) . . . . . . . . . . . . . . . . . . . 54
4.3
Solution of D.Es. by variation of parameters method . . . . . . . . 59
4.4
Solution of D.Es. by undetermined coefficients method . . . . . . . 62
4.5
Euler’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4.5.1 4.5.2
Second order D.E. of the form Second order D.E. of the form
5 Series Solution
d2 y dx2 d2 y dx2
= f (y) . . . . . . . . . 68 dy = f (y, dx ) . . . . . . . . 68 74
5.1
Series solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5.2
Singular points, the method of Frobenius . . . . . . . . . . . . . . 82
5.3
Bessel’s Equation
. . . . . . . . . . . . . . . . . . . . . . . . . . 89
6 Applications of Differential Equations
96
6.1 6.2
Motion of a particle in a resistive medium . . . . . . . . . . . . . . 96 Schr¨odinger Equation . . . . . . . . . . . . . . . . . . . . . . . . 98
6.3
Electric circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
6.4
Table of Integration . . . . . . . . . . . . . . . . . . . . . . . . . 105
6.5 6.6
Answers to exercises . . . . . . . . . . . . . . . . . . . . . . . . . 109 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
1
Chapter 1 Differential Equations 1.1
The origin of differential equations
A differential equation (D.E) is an equation that involves derivatives. (i)
dy d2 y dy d2 y 2 dy = x + 5, (ii) + 3 + 2y = 0, (iii) ( ) + + y = x3 , dx dx2 dx dx2 dx (iv)
∂z ∂z ∂ 2z ∂ 2z =z+ , (v) + = x2 + y. ∂x ∂y ∂x2 ∂y 2
If the equation contains a single independent variable, as in (i), (ii) and (iii) the derivatives are ordinary derivatives and the equation is called an ordinary differential equation. If the equation contains two or more independent variables, as in (iv) and (v), the derivatives are partial derivatives and the equation is called partial differential equation.
1.1.1
The order of a D.E.
Is the order of the highest derivative that appears in the equation. Thus (i) and (iv) are first order while (ii), (iii) and (v) are second order.
1
1.1.2
The degree of a D.E
Is the degree of the highest ordered derivative that appears in the equation. Thus (i), (ii) (iv), and (v) are first degree while (iii) is second degree. The origin of a D.E could be (a) geometric, (b) physical, (c) primitive. A primitive is a relation between the variables which involves n arbitrary constants, e.g. y = Ax2 + Bx + C. In general a primitive involving n arbitrary constants will give rise to a D.E., of order n, free of arbitrary constants. A geometrical origin: dy A slope of a curve is dx and the equation of the straight line is y = mx + c, where m=
dy , dx
c the intercept. Hence, one can write y =
dy x dx
− 2, as an equation of a
straight line. This equation represents a first order D.E. Physical origin A particle of mass m moves vertically downwards due to gravity. If the air resistance is −kv, k= constant. Find its equation of motion. From Newton’s second law of motion we have ma = mg − kv, a =
dy d2 y , v= , 2 dt dt
where g is the acceleration due to gravity. This can be written as d2 y k dy =g− . 2 dt m dt This is a second order D.E. of degree 1. Primitive (i) Given y = Ax2 + Bx + C, where A, B, C are arbitrary constants. Set up the D.E. for this primitive. d2 y dy = 2Ax + B, = 2A, dx dx2 2
d3 y = 0. dx3
Thus
d3 y = 0, dx3
is free of any arbitrary constants. (ii) Find the D.E. associated with x2 y + y 2 x3 = 1. Differentiate both sides with respect to (w.r.t) x, we get 2xy + x2 which can be written as
dy dy + 2yx3 + 3x2 y 2 = 0, dx dx
dy −3x2 y 2 − 2xy = , dx x2 + 2yx3
which is a D.E. of first order. (iii) Find the D.E. associated with the primitive y = A cos ax + B sin ax, where a, A, B are arbitrary constants. dy d2 y = −Aa sin ax + Ba cos ax, = −Aa2 cos ax − Ba2 sin ax, 2 dx dx d2 y = −a2 (A cos ax + B sin ax) = −a2 y. 2 dx Hence,
d2 y + a2 y = 0, dx2
which is a D.E. of second order. (iv) Obtain the D.E. associated with the primitive y = Ae2x + Bex + C, where A, B, C are arbitrary constants. d2 y d3 y dy 2x x = 2Ae2x + Bex , = 4Ae + Be , = 8Ae2x + Bex . dx dx2 dx3 3
To eliminate the constant, we write dy d3 y d2 y dy − 3 = 4Ae2x , − = 2Ae2x , dx dx dx2 dx and hence,
dy d3 y d2 y dy − 3 = 4Ae2x = 2( 2 − ), dx dx dx dx
or
d3 y d2 y dy − 3 + 2 = 0, dx3 dx2 dx which is the D.E. that corresponds to the primitive above and is free of any arbitrary constant. (v) Find the D.E. associated with the primitive y = Cx2 + C 2 . dy 1 dy = 2Cx, C = , dx 2x dx and hence, y=
1 dy 2 1 dy 2 x +( ), 2x dx 2x dx
or
dy 2 dy ) + 2x3 − 4x2 y = 0, dx dx which is first order and second degree. (
(vi) Find the D.E. associated with the family of circles of radius r centered on the x-axis. This can be written as y 2 + (x − C)2 = r2 , where C is an arbitrary constant. Differentiating both sides, we obtain 2y
dy + 2(x − C) = 0, dx
Therefore, y2 + y2(
or (x − C) = −y dy 2 ) = r2 , dx 4
dy . dx
is our required D.E. Exercise 1.1 (1) Find the order and the degree of the following D.Es. dy + 3xy = 0 , dxq dy d2 y (iii) + y = dx 2 dx
(i)
2
(ii) 3 ddt2z + 4 dz = z + t. dt 2
dy 2 d y ) y + dx , (iv) ( dx 2 + x = 0. (2) Find the D.Es. associated with the following primitives
(i) y = Ax , (ii) y = sin(x + A) , (iii) y = Ax + B, (iv) y = Ax2 + Bx , (v) y = Aex + Bx, where A, B are arbitrary constants. (3) Find the D.E. of all straight lines at a unit distance from the origin. (4) What are the D.Es. for the following primitives (i) y = c1 e2x + c2 e−3x , (ii) y = cx + 2x2 , (iii) y = c1 sin x + c2 cos x.
1.1.3
The solution of differential equations
This is that of finding a relation between the variables involving n independent arbitrary constants, together with the derivatives obtained from it, satisfying the differential equation. Example. d3 y dx3
dy 2 = 0, corresponds to a primitive y = Ax2 + Bx + C and y 2 ( dx ) + y 2 = r2 ,
that corresponds to the primitive y 2 + (x − C)2 = r2 . Existence Theorem A D.E. of the form y 0 ≡
dy dx
= g(x, y) for which
(a) g(x, y) is continuous and single valued over a region R of point (x, y) (b)
∂g ∂x
exists and continuous at all points in R, admits many solutions f (x, y, C) =
0, where C is an arbitrary constant, such that through each point of R there passes one and only one curve of the family f (x, y, C) = 0.
5
1.2
The general and particular solutions of the D.Es.
The primitive is some times called the general solution of a D.E and the particular solution is obtained from the general solution by assigning definite values to the arbitrary constants appearing in the primitive. These are some times referred to as initial conditions. Example.
d3 y = 0, dx3 has the general solution(primitive) y = Ax2 + Bx + C. If we assign certain values for the constants A, B, C we obtain the particular solutions. Example. Set A = 1, B = −1, C = 3 then y = x2 − x + 3 is the corresponding particular solution. If we set A = 0, B = 1, C = −5, we obtain the particular solution, y = x − 5, and so forth. If the particular solution can not be obtained from the general solution by certain choice of the arbitrary constant, the solutions are called singular solutions Examples. (i) Show that y = 2x + Cex
(1.1)
is the general solution of the D.E. dy − y = 2(1 − x), dx and find the particular solution satisfied by x = 0, y = 3. Substituting in the above D.E. one obtains
dy dx
= 2+Cex
(2 + Cex ) − (2x + Cex ) = 2 − 2x = 2(1 − x) = R.H.S. Therefore, y = 2x+Cex is the general solution. The particular solution is obtained by finding the value of C. The point x = 0, y = 3 in equation (1.1) gives 3 = 0+C, 6
or C = 3. Hence, y = 2x + 3ex is the required particular solution. (ii) Show that y = C1 ex + C2 e2x + x,
(1.2)
is the general solution (primitive) of the D.E. d2 y dy − 3 + 2y = 2x − 3, 2 dx dx and find the equation of the curve through the points (0,0) and (1,0). dy d2 y = C1 ex + 2C2 e2x + 1, 2 = C1 ex + 4C2 e2x , dx dx hence, d2 y dy − 3 + 2y = C1 ex + 4C2 e2x − 3(C1 ex + 2C2 e2x + 1) + 2(C1 ex + C2 e2x + x) 2 dx dx = 2x − 3 = R.H.S. The points (0,0) and (1,0) give, in equation (1.2), the two equations: and 0 = C1 e + C2 e2 + 1,
0 = C1 + C2
which can be solved to yield C1 = −C2 =
1 , e2 −e
thus the equation of the curve
through the given points is y=
ex e2x − +x e2 − e e2 − e
which is a particular solution. (iii) Show that (y − C)2 = Cx, is the general solution of the D.E. 4x(
dy dy 2 ) + 2x − y = 0, dx dx 7
(1.3)
and find the equation of the curves through the point (1,2). Differentiating (1.3) yields dy 2(y − C) = C, dx hence, the D.E. becomes 4x(
C C 4xC 2 + 4xC(y − C) − 4y(y − C)2 )2 + 2x( )−y = =0 2(y − C) 2(y − C) 4(y − C)2
=R.H.S. The point x = 1, y = 2 gives, in equation (1.3), (2 − C)2 = C, or 4 − 4C + C 2 = C, C 2 − 5C + 4 = 0, which can be written as (C − 1)(C − 4) = 0, C = 1, C = 4. This gives two particular solutions, viz., (y − 1)2 = x and (y − 4)2 = 4x. (iv) If the general solution of the D.E., dy y = dx x
is y = Cx,
(1.4)
where C is an arbitrary constant. Find the equation of the curve through the points (a) (1,2), (b) (0,0). The point (1,2) gives, in equation (1.4), C = 2 so that the equation of the curve is y = 2x. The point (0,0) gives, in equation (1.4), 0=0, and C is not determined!, and all of the curves that pass through the point (0,0). This is because g(x, y) =
y x
is not continuous at the point (origin) and hence, the existence theorem assures that one and only one curve of the family y = Cx passes through each point except the origin. (v) If xy = C(x − 1)(y − 1), 8
(1.5)
obtain the D.E.
dy + y(y − 1) = 0, dx we see that the solution y = 0 can be obtained from the general solution equation x(x − 1)
(1.5) by putting C = 0, but y = 1 can not be obtained from the general solution of equation (1.5) by any value of C and hence, this solution is a singular solution. (vi) The primitive y = Cx + 2C 2 , and
(1.6)
dy = C, dx
give y=
dy dy x + 2( )2 . dx dx
(1.6a)
The equation of the parabola y = − 18 x2 satisfies equation (1.6a). We see that the primitive is represented by a family of straight lines (1.6) and it is clear that equation of a parabola can’t be obtained from this by any value of C, such a solution is called singular solution. (vii) Show that the two general solutions y = C1 cos x + C2 sin x ,
(1.7)
y = A sin(x + B) ,
(1.8)
and where A, B, C1 , C2 are arbitrary constants, satisfies the D.E. d2 y + y = 0, dx2 and show that how they are made equivalent. From equation (1.7) we have, d2 y dy = −C1 sin x + C2 cos x, 2 = −C1 cos x − C2 sin x = −(C1 cos x + C2 sin x), dx dx 9
or
d2 y +y =0 . dx2
Now consider equation (1.8): dy d2 y = A cos(x + B), = −A sin(x + B) = −y, dx dx2 and
d2 y +y =0 . dx2 Now equation (1.8) can be written as y = A sin(x + B) = A sin B cos x + A cos B sin x, and by setting C1 = A sin B, C2 = A cos B, the above equation becomes y = C1 cos x + C2 sin x, which is the same as (1.7) above. (viii) Show that ln x2 + ln
y2 = A + x, x2
can be written as y 2 = Bex , where B is an arbitrary constant. ln x2 + ln
2 y2 2y = ln(x ) = ln y 2 = A + x, x2 x2
or y 2 = eA+x = eA .ex = Bex , where B = eA =constant. Exercise 1.2 Which of the following solutions are general solutions and which are particular solutions 10
dy for the D.E. x dx = 2y.
(i) y = 2x2 ,
dy for the D.E. y dx + x = 0.
(ii) x2 + y 2 = C, (iii) y = Cx + C 4 ,
dy dy 4 for the D.E. y = x dx + ( dx ). dy for the D.E. 2x3 dx = y(y 2 + 3x2 ).
(iv) (1 − x)y 2 = x3 , (v) y = C1 ex + C2 e−x ,
for the D.E.
d2 y dx2
− y = 0.
(vi) y = C1 ex + C2 e2x ,
for the D.E.
d2 y dx2
dy − 3 dx + 2y = 0.
1.2.1
First order differential equations
A D.E. of first order and first degree can be written in the form M (x, y)dx + N (x, y)dy = 0.
(1.9)
Example. dy 2x − y = , dx y + 2xy + 3 can be written as (2x − y)dx + (y + 2xy + 3)dy = 0, where M (x, y) = 2x − y ,
N (x, y) = y + 2xy + 3 .
If M dx + N dy is a complete (exact) differential of a function µ(x, y), i.e., M dx + N dy = dµ, then µ = C is the primitive (general solution) of the D.E. Example. 3x2 y 2 dx + 2x3 ydy = 0,
11
is an exact D.E., since 3x2 y 2 dx + 2x3 ydy = d(x3 y 2 ). Its primitive is µ = x3 y 2 = C
or y 2 = Cx−3 .
If equation (1.9) is not exact but ξ(x, y)(M dx + N dy) = dµ, where ξ(x, y) is known as an integrating factor (I.F.), then µ = C is the primitive of the D.E. Example. 3ydx + +2xdy = 0, is not an exact differential equation, but when multiplied by ξ = x2 y becomes exact. Thus x2 y(3ydx + 2xdy) = 3x2 y 2 dx + 2x3 ydy = d(x3 y 2 ), is exact. Its primitive is µ = x3 y 2 = C, or y 2 = Cx−3 . If equation (1.9) is not exact and no integrating factor can be found readily, it may be possible by a change of variables to obtain an equation for which the integrating factor can be found. Example. Solve (x + y + 1)dx + (2x + 2y + 3)dy = 0. Now let x + y = t and dx + dy = dt, hence, (t + 1)(dt − dy) + (2t + 3)dy = 0 , or dy + which can be written as dy + dt −
1 dt = 0, t+2 12
t+1 dt = 0, t+2
by integrating this we obtain, y + t − ln(t + 2) = C , or 2y + x − ln(x + y + 2) = C, which is the primitive. Exercise. Try the solution by putting x + y + 1 = t. A D.E. for which the integrating factor can be found readily has the form f1 (x)g2 (y)dx + f2 (x)g1 (y)dy = 0, which when multiplied by (integrating factor) 1 , f2 (x)g2 (y) reduces to
f1 (x) g1 (y) dx + dy = 0, f2 (x) g2 (y)
and finally
Z
Z f1 (x) g1 (y) dx + dy = C, f2 (x) g2 (y)
is the primitive. Such equation is called variables separable. Examples. (i) Find the primitive for (3x2 y − yx)dx + (2x3 y 2 + x3 y 4 )dy = 0. y(3x2 − x)dx + x3 (2y 2 + y 4 )dy = 0, in this case f1 (x) = 3x2 − x, f2 (x) = x3 , g2 (y) = y, g1 (y) = 2y 2 + y 4 . Hence,
Z
Z 2y 2 + y 4 3x2 − x dx + dy = C, x3 y
13
and
Z
Z 3 1 ( − 2 )dx + (2y + y 3 )dy = C, x x y4 3 ln x + x−1 + y 2 + = C, 4 is the required primitive.
(ii) Find the general solution of the D.E. (x − 1)2 ydx + x2 (y + 1)dy = 0. Separating the variables (x and y) one can write the above equation as (x − 1)2 y+1 dx + dy = 0. 2 x y Integrate both sides to obtain Z
(1 −
Z 2 1 1 + 2 )dx + (1 + )dy = C, x x y
or x − 2 ln x −
1 + y + ln y = C, x
is the primitive. If equation (1.9) admits a solution f (x, y, C) = 0, where C is an arbitrary constant, there exist many integrating factors, ξ(x, y) such that ξ(x, y)(M (x, y)dx + N (x, y)dy) = 0, is exact. Example. xdy − ydx = 0, can be put as
dy dx − = 0, y x 14
and integrate to obtain y y ln y − ln x = ln C, ln( ) = ln C, = C, x x and y = Cx, is the primitive. (b) If we multiply by
1 , x2
we get
xdy − ydx y = d( ) = 0 i.e., y = C1 x. 2 x x Similarly, if we multiply the equation by we get
1 , y2
xdy − ydx x = d(− ) = 0, 2 y y
or
−x y 1 = C2 , =− = C = C1 . y x C2 Thus, the form of the primitive is not unique.
1.2.2
Homogeneous Equations
A function f (x, y) is called homogeneous function of degree n if f (λx, λy) = λn f (x, y). Examples. (i) f (x, y) = x3 − xy 2 , f (λx, λy) = (λx)3 − (λx)(λy)2 = λ3 (x3 − xy 2 ) = λ3 f (x, y), hence, the function is homogeneous of degree 3. (ii) The function f (x, y) = x2 + y, 15
is not homogeneous, since f (λx, λy) = (λx)2 + (λy) = λ(λx2 + y) 6= λn f (x, y), for some number n. (iii)
1 , + 2x2 y + 5y 3 1 f (λx, λy) = , (λx)3 + 2(λx)2 (λy) + 5(λy)3 1 f (λx, λy) = 3 3 = λ−3 f (x, y), λ (x + 2x2 y + 5y 3 ) f (x, y) =
x3
hence, the equation is homogeneous of degree -3. A D.E. M dx + N dy = 0, is called homogeneous if both M and N are homogeneous and of the same degree. A homogeneous D.E. can be made integrable by the transformation y = vx Examples. (i) Find the general solution of the D.E. (x3 + y 3 )dx − 3xy 2 dy = 0. It is clear this equation is a homogeneous equation of degree 3 (check this). Therefore one can use the transformation y = vx, so that dy = vdx + xdv, and hence, our equation becomes (x3 + v 3 x3 )dx − 3x(vx)2 (vdx + xdv) = x3 (1 − 2v 3 )dx − 3v 2 x4 dv = 0, or
dx 3v 2 dv − dv = 0. x 1 − 2v 3
Integrating this we obtain ln x + ln (1 − 2v 3 )1/2 = ln(x(1 − 2v 3 )1/2 ) = ln C, 16
or
y3 ) = C 2 , x3 − 2y 3 = C 2 x. x3 (ii) Find the general solution of the D.E. x2 (1 − 2
(x2 − xy)dx + x2 )dy = 0. Clearly, the equation is homogeneous of degree 2 (both M and N are homogeneous of degree 2), therefore one uses the substitution y = vx so that dy = vdx + xdv, and hence [x2 − x(vx)]dx + 2x2 [vdx + xdv] = 0, or
Z 2
3
x (1 − v + 2v]dx + 2x dv = 0,
Z dx dv = −2 . x 1+v
This yields ln x + 2 ln(1 + v) = ln C, Thus, the general solution is y=
√
ln x(1 + v)2 = ln C.
Cx − x.
Suppose
dy y = f( ) dx x Such a D.E. is called homogeneous equation. In such a case, we make the following transformation xy = v, so that x
dv + v = f (v). dx
Example. Find the general solution of the following D.E. below. dy y y = 2 + ( )2 . dx x x Let
y x
= v, so that this equation becomes x
dv + v = 2v + v 2 , dx 17
or x
dv = v + v2, dx
which is a separable equation, i.e., Z
Z Z dx dv dv = = , 2 v+v v(1 + v) x
or Z
Z Z 1 1 dx Z dv dv dx ( − )dv = , ( − )= , ln v−ln(1+v) = ln x+ln C. v 1+v x v 1+v x Thus, y v y Cx2 x = Cx, or = = Cx, y = . 1+v 1 + xy x+y 1 − Cx Exercise 1.3
(1) Find the general solution (primitive) of the following D.Es. (i) xydx + (1 + x2 )dy = 0 , (ii) (y 2 − x2 )dx + xydy = 0 , (iii) (x3 − 3y 3 )dx + 3xy 2 dy = 0 , (iv) (xy + y 2 )dx + (x2 − yx)dy = 0. (2) Find the solution of the following D.Es. by suitable substitution (i) (y − xy 2 )dx − x(1 + xy)dy = 0 , (ii) dy − (y − 4x)2 dx = 0. (3) Find the particular solution of the following D.Es. (i) (y 2 + xy)dx − x2 dy = 0, (ii) xdy + 2ydx = 0,
when x = 1, y = 1,
when x = 2, y = 1,
(iii) (y 2 + x2 )dx + xydy = 0,
when x = 1, y = −1.
(4) Solve the following D.E. (x2 + y 2 )dx − xydy = 0. 18
(5) Solve the following D.E. dy = 0 ; y(1) = 1. dx (6) Find all functions f (x) such that the D.E. 2xy 3 + 3x2 y 2
y 2 sin x + yf (x)
dy = 0, dx
is exact.
Solve the D.E. for these functions f (x). (7) The D.E.
dy + x2 + y = 0 dx is known to have an integrating factor ξ = x. Find all possible functions f (x). f (x)
1.2.3
Equations reducible to homogeneous equation
Consider an equation of the type dy a1 x + b1 y + c1 = , dx a2 x + b2 y + c2 where a1 , a2 , b1 , b2 , c1 , c2 are constants. When c1 = 0 = c2 or a1 b2 − a2 b1 = 0 the solution of the equation has been investigated previously. However, when a1 b2 − a2 b1 6= 0 one employs a different technique. Example. Solve the following D.E.
dy 2x + 3y − 4 = . dx 4x + y − 3 Let 2X + 3Y = 2x + 3y − 4 , 4X + Y = 4x + y − 3. Solving for X, Y gives X =x− Since
dy dx
=
dY , dX
1 , 2
Y =y−1 .
the D.E. becomes dY 2X + 3Y = . dX 4X + Y 19
(1.10)
This is a homogeneous equation in X and Y . Its solution can easily be found by putting Y = V X so that dY = V dX + XdV . Thus, Z Z dV 2X + 3V x 2 + 3V dX 4+V V +X = = , or =− dV. 2 dX 4X + V X 4+V X V +V −2
Thus, ln X = A + Now V =
Y X
=
y−1 x− 12
2 5 ln(2 + V ) − ln(1 − V ). 3 3
so that
1 2 2x + y − 2 5 x − y + 12 ln(x − ) = A + ln − ln . 2 3 3 x − 12 x − 12 Exercise 1.4 Solve the following first order D.Es. (i) (x + 2y − 3) dx − (2x + y − 3) dy = 0 , (ii) (x − 8y + 7) dy − (x − y) dx = 0 . (iii) (3x+2y−3)dx−(2x+3y−2)dy = 0 , (iv) (y+x+1)dx−(2y+2x+1)dy = 0 . (v) (x − 2y + 3)dx − (x − 2y + 5)dy = 0 , (vi) (x + y)dx − (x + y − 1)dy = 0 .
20
Chapter 2 Exact Differential Equations 2.0.4
Exact differential equation
The necessary and sufficient condition for a D.E. of the form M (x, y)dx + N (x, y)dy = 0, to be exact is that
∂M ∂N = . ∂y ∂x
is the derivative of M w.r.t y keeping x constant and ( ∂M ∂y N w.r.t x keeping y constant). Examples. (i) (x2 − y)dx + (y 2 − x)dy = 0, is exact since
Therefore,
(2.1)
∂M ∂N = −1, = −1. ∂y ∂x ∂N ∂M = . ∂y ∂x
(ii) (y 2 − x)dx + (x2 − y)dy = 0, 21
∂N ∂x
is the derivative of
is not exact since
∂M = 2y, ∂y
∂N = 2x, ∂x
and
∂M ∂N 6= . ∂y ∂x If equation (2.1) is exact then it is possible to find µ = µ(x, y) such that M dx + N dy = dµ,
(2.2)
but
∂µ ∂µ dx + dy, (2.3) ∂x ∂y (this is the definition of the total derivative of a function of two variables). Comdµ =
paring (2.2) and (2.3) one obtains M= This gives µ= or
Z x
Z x
∂µ ∂x
and N =
M dx + φ(y),
∂µ , ∂y
and N =
∂µ , ∂y
∂µ . ∂x From these two equations one can find the form of µ(x, y). R R (Here x ( y ) means that in the integration y(x) is treated as a constant). If the µ=
N dy + φ(x),
and M =
equation is not exact then an integrating factor ξ must be sought. (a) if ∂M ∂y
−
∂N ∂x
N
= f (x), R
a function of x only, then the integrating factor ξ(x, y) = e Example. Solve the following D.E. (x2 + y 2 + x)dx + xydy = 0, 22
f (x)dx
.
∂M ∂N = 2y , = y. ∂y ∂x The equation is not exact and we must look for an integrating factor. Now ∂M ∂y
−
∂N ∂x
N
2y − y 1 = = f (x), xy x
=
hence,
R
ξ=e
1 dx x
= eln x = x.
Hence, x[(x2 + y 2 + x)dx + xydy] = (x3 + xy 2 + x2 )dx + x2 ydy = 0, is exact. (b) If ∂M ∂y
−
∂N ∂x
M
= −g(y), R
a function of y only, then the integrating factor ξ(x, y) = e
g(y)dy
.
Example. Solve (2xy 4 ey + 2x3 y 2 + y)dx + (x2 y 4 ey − x4 y − 3x)dy = 0, ∂M ∂N = 8xy 3 ey + 2xy 4 ey + 4x3 y + 1, = 2xy 4 ey − 4x3 y − 3, ∂y ∂x the equation is not exact and we seek an integrating factor (ξ). Note that ∂M ∂y
− M
∂N ∂x
=
4 = −g(y), y R
hence, the integrating factor ξ(x, y) = e
g(y)dy
R
=e
− y4 dy
= e−4 ln y = eln y
−4
y . Therefore, y −4 [(2xy 4 ey + 2x3 y 2 + y)dx + (x2 y 4 ey − x4 y − 3x)dy] = (2xey + 2x3 y −2 + y −3 )dx + (x2 ey − x4 y −3 − 3xy −4 )dy = 0, 23
−4
=
is exact. (c) If the equation is homogeneous then the integrating factor is ξ=
1 . Mx + Ny
Example. Solve (x4 + y 4 )dx − xy 3 dy = 0, the equation is homogeneous of degree 4 (check this) and the integrating factor ξ=
1 1 1 = 4 = 5. 4 3 Mx + Ny (x + y )x + (−xy )y x
Hence, ξ= so that
1 , x5
1 4 1 y4 y3 4 3 [x + y )dx − xy dy] = ( + )dx − dy = 0, x5 x x5 x4
is exact.
2.1
Solution of an Exact D.E.
Examples. (i) Find the primitive of the D.E. (2x3 + 3y)dx + (3x + y − 1)dx = 0. Now
∂M = 3, ∂y
∂N = 3, ∂x
and therefore, the equation is exact. Set µ=
Z x
M dx + φ(y), 24
and
N=
∂µ . ∂y
Therefore, µ=
Z x
(2x3 + 3y)dx + φ(y) =
and 3x + y − 1 = or
x4 + 3yx + φ(y) , 2
∂µ dφ = 3x + , ∂y dy
dφ = y − 1. dy
Integrating this, we obtain φ =
y2 2
− y and the primitive µ = C implies 21 x4 +
3yx + 12 y 2 − y = C. (ii) Find the primitive of (4x3 y 3 − 2xy)dx + (3x4 y 2 − x2 )dy = 0. ∂M ∂N = 12x3 y 2 − 2x, = 12x3 y 2 − 2x, ∂y ∂x and therefore, the equation is exact. Set µ= hence, µ=
Z x
Z x
M dx + φ(y) and N =
∂µ , ∂y
(4x3 y 3 − 2xy)dx + φ(y) = x4 y 3 − x2 y + φ(y),
and 3x4 y 2 − x2 =
∂µ dφ = 3x4 y 2 − x2 + , ∂y dy
dφ = 0, φ = const., dy
hence, µ = x4 y 3 − x2 y + const. and the primitive is x4 y 3 − x2 y = C. Find the constant a such that the following D.E. is exact, x + ye2xy + axe2xy 25
dy = 0. dx
We have here, M = x + ye2xy , N = axe2xy , so that ∂M ∂N = 2yxe2xy + e2xy , = 2ayxe2xy + ae2xy . ∂y ∂x The exact equation requires
∂M ∂y
=
∂N ∂x
and therefore a = 1.
Find the general solution of the D.E. 1 2 dy y + 2yex + (y + ex ) = 0. 2 dx We see that M = 12 y 2 + 2yex , N = y + ex so that However,
∂M ∂y
− ∂N ∂x N
=
y+ex y+ex
∂M ∂y
= y + 2ex ,
∂N ∂x
= ex .
= 1. Hence, the integrating factor ξ = ex . Thus, the
exact D.E. becomes 1 dy 1 dy ex [ y 2 + 2yex + (y + ex ) ] = y 2 ex + 2ye2x + (yex + e2x ) = 0. 2 dx 2 dx Therefore, one can write µ=
Z x
M dx + φ(y) ,
N=
∂µ . ∂y
Therefore, µ=
Z x 1 1 ( y 2 ex + 2ye2x )dx + φ(y) = y 2 ex + ye2x + φ ,
2
2
yex + e2x = yex + e2x +
dφ , dy
This gives φ0 = 0, or φ = const. hence, µ = 12 y 2 ex + 2ye2x = C. Find the general solution of the following D.E. (3xy + y 2 ) + (3xy + x2 )
dy = 0. dx
It is evident that the equation is homogeneous, so that we seek the solution with an integrating factor ξ = ξ=
(3xy +
y 2 )x
1 . M x+N y
Therefore,
1 1 1 = = . 2 2 2 + (3xy + x )y 4xy + 4x y 4xy(x + y) 26
Thus the exact D.E. will be 3x + y 3y + x dy + = 0. 4x(x + y) 4y(x + y) dx This gives µ = xy(x + y)2 = C. This D.E. can also be solved by putting y = vx as is done before. Find the general solution of the following D.E. (x + 3y 2 )dx + 2xydy = 0. ∂M ∂N = 6y , = 2y. ∂y ∂x It clear that the equation is not exact and we must look for an integrating factor. We see that,
∂M ∂y
−
∂N ∂x
N
= R
Hence, the integrating factor ξ = e
6y − 2y 2 = = f (x). 2xy x 2 dx x
2
= eln x = x2 . Thus the exact D.E. is
(x3 + 3y 2 x2 )dx + 2x3 ydy = 0. Therefore, µ=
Z x
M dx =
Z x
1 ∂µ dφ (x3 +3y 2 x2 )dx+φ(y) = x4 +y 2 x3 +φ(y) , 2yx3 = 2x3 y+ , 4 ∂y dy
hence, φ0 = 0, φ = const.. Thus, 1 µ = x4 + y 2 x3 = C. 4 Exercise 2.1 Find the primitive of the following D.Es. (1) (i) (x2 − y)dx − xdy = 0, (iii) y(x − 2y)dx − x2 dy, (2) The D.E.
(ii) (x4 + y 4 )dx − xy 3 dy = 0,
(iv) (2y − x3 )dx + xdy. ex sec y − tan y + 27
dy =0 dx
has an integrating factor of the form e−ax cos y for some constant a. Find a, and then solve the D.E. (3) Determine the constant a so that the following D.E. is exact and solve the resulting equation
1 1 (ax + 1) dy + 2+ = 0. 2 x y y3 dx
(4) Find the general solution of the following D.Es. (y 2 − 3x2 )ydy − (x2 − 3y 2 )xdx = 0. (5) Prove that (3x2 y 2 + cos xy − xy sin xy) dx + (2x3 y − x2 sin xy) dy = 0, is an exact D.E., and obtain its general solution.
28
Chapter 3 First Order Linear Differential Equations 3.0.1
Linear Differential Equations
The equation
dy + P (x)y = Q(x) , (3.1) dx whose left hand side (LHS) is linear in both the dependent and its derivatives, is called a linear differential equation. Example. dy + 3xy = sin x, dx is a linear D.E., but
dy + 3xy 2 = sin x, dx R
is not. Multiply (3.1) by the integrating factor e
P (x)dx
, to get
R R R dy R P (x)dx d e + yp(x)e P (x)dx = (ye P (x)dx ) = Q(x)e P (x)dx . dx dx
Then
R
ye
Z P (x)dx
=
R
Q(x)e 29
P (x)dx
dx,
so that −
R
y=e
Z P (x)dx
R
[ Q(x)e
P (x)dx
dx + C],
is the primitive (general solution). Equivalently, we substitute y = uv in equation (3.1) to obtain du dv v + u + P (x)uv = Q. dx dx Or dv du u( + P v) + v = Q. dx dx Now choose dv + Pv = 0 dx which gives R v = e−
P dx
.
The above equation yields v du = Q, which gives dx Z
u=
Q dx + C. v
Hence, −
y = uv = e
R
Z P dx
[
Z R Q Q − P dx R dx + C], dx + C] = e [ − v e P dx
Finally, one writes y = e−
R
Z P (x)dx
R
[ Q(x)e
P (x)dx
dx + C]
to obtain the primitive, which is the same as equation (3.1).
3.0.2
Bernoulli’s Equation
This is an equation of the form dy + P (x)y = Q(x)y n , dx 30
(3.2)
which can be reduced to dz + (1 − n)P (x)z = (1 − n)Q(x), n 6= 1 dx by the transformation z = y 1−n . Dividing equation (3.2) by y n we obtain y −n
dy + P (x)y 1−n = Q(x), dx
dy dy dz = , dx dz dx
so that
dy 1 dz dy 1 dz = y n , y −n = dx 1 − n dx dx 1 − n dx and hence, equation (3.2) becomes dz + (1 − n)P (x)z = (1 − n)Q(x). dx Which is a linear D.E. that can easily be solved. Examples. (i) Solve
dy + 2xy = 4x, dx
Z
Z
P (x)dx = and hence, y = e− yields −x2
y=e
2xdx = x2 ,
Z
R
P (x)dx
Z
R
( Q(x)e
P (x)dx
dx + C),
2
2
[ 4xex dx + C] = 2 + Ce−x , (Q = 4x)
is the required primitive. (ii) Solve the following D.E. x
dy = y + x3 + 3x2 − 2x. dx
This can be put in the form dy 1 − y = x2 + 3x − 2, dx x 31
Z
Z
1 1 dx = − ln x = ln , x x 1 1 = e− ln x = eln x = , x
P (x)dx = − R
P (x)dx
e and hence, y = e−
R
Z P (x)dx
R
[ Q(x)e
P (x)dx
Z 1 dx + C] = x[ (x2 + 3x − 2) dx + C] x
1 y = x3 + 3x2 − 2x ln x + Cx. 2 (iii) Solve the following D.E. (x − 2)
dy = y + 2(x − 2)3 . dx
This can be put in the form dy 1 − y = 2(x − 2)2 , dx x − 2 Z
Z
P (x)dx = and
R
e hence,
−1 dx = − ln(x − 2) = ln(x − 2)−1 , x−2 P (x)dx
= eln(x−2)
−1
Z
y = (x − 2)[ 2(x − 2)2
=
1 , x−2
1 dx + C], x−2
or y = (x − 2)3 + C(x − 2), is the required primitive. (iv) Solve
dy − y = xy 5 . dx This is a Bernoulli type equation and we therefore use the transformation z = y 1−n = y 1−5 = y −4 , 32
and the equation becomes dz + (1 − n)P z = (1 − n)Q, dx dz − 4z = −4x , dx This has the usual solution −
z=e
R
Z
P (x)dx
R
[ Q(x)e
P (x)dx
and e− and
(Q = x, n = 5). Z
dx + C],
R
P (x)dx
Z
P (x)dx = −
4dx = −4x,
= e4x ,
Z
z = e4x [ −4xe−4x dx + C]. Writing the y dependence, we find Z
y −4 = e4x [
−4xe−4x dx + C] = x +
1 + Ce4x . 4
(v) Solve the following D.E., dy + (tan x)y = y 3 sec4 x. dx Putting z = y −2 we find dz − 2(tan x)z = −2 sec4 x. dx R R exp[ P (x)dx] = exp[ (−2 tan x)dx] = cos2 x, so that z cos2 x = C − 2 tan x, or y 2 =
cos2 x . C − 2 tan x
Exercise 3.1 Find the general solution of the following D.Es. (i) (iii)
dy + 2xy + xy 4 = 0, dx
(ii)
dy 1 + y = sin x, dx x
dy dy = y tan x − sec x, (iv) + y tan x = y 2 sec3 x. dx dx 33
3.0.3
First order D.Es. with higher degrees
A D.E. of first order has the form f (x, y, Upon writing, p =
dy dx
dy ) = 0. dx
this becomes f (x, y, p) = 0. If the degree of p is greater
than one, the equation is first order but higher degree. The general form is pn + P1 (x, y)pn−1 + P2 (x, y)pn−2 + ... + Pn (x, y)y = 0, where P1 (x, ), P2 (x, y), ...., Pn (x, y) are functions of x and y. We discuss the following situations: (1) Equation solvable for p: In this case we will have (p − F1 )(p − F2 )(p − F3 ).....(p − Fn ) = 0, where all F’s are functions of x and y, and the solution is p = F1 (x, y),
p = F2 (x, y), ...
which can be solved to obtain f1 (x, y, C1 ) = 0,
f2 (x, y, C2 ) = 0, ...
and the primitive is the product f1 (x, y, C1 ).f2 (x, y, C2 ).... = 0. Example. Solve p2 − 3xp + 2x2 = 0. This equation can be written as (p − x)(p − 2x) = 0, 34
and the solutions are p = x, p = 2x, i.e., Z dy x2 1 p = x, = x, y = xdx = + C1 , y − x2 − C1 = f1 (x, y, C1 ) = 0 , dx 2 2
and Z dy p = 2x, = 2x, y = 2 xdx = x2 + C2 , y − x2 − C2 = f2 (x, y, C2 ) = 0 , dx
where C1 , C2 are constants. Hence, the general solution is (y −
x2 − C1 )(y − x2 + C2 ) = 0. 2
(2) Equation solvable for x: i.e., x = f (y, p), differentiate both sides w.r.t. y we obtain dx ∂f ∂f dp dp = + ≡ F (y, p, ) dy ∂y ∂p dy dy and we solve for
1 dp = F (y, p, ), p dy
to obtain φ(y, p, C) = 0 Example. p3 − 2xyp + 4y 2 = 0. This can be written as 2x =
p2 4y + , y p
and differentiate w.r.t. y to get dp dp − p2 4p − 4y dy 2yp dy 2 = + , p y2 p2
35
which can be written as (p − 2y
dp )(2y 2 − p3 ) = 0, dy
that has the solution p − 2y
dp = 0, p2 = Ky dy
and 2y 2 = p3 .
Substituting the first solution in the main D.E. equation, one obtains 4y . 2x = K + √ Ky This can be written as (2x − K)2 = 16
y K
or 2y = C(C − x)2 ,
where 2C = K.
The second solution (2y 2 = p3 ) when substituted in the original equation gives = dy , which gives y = Cx3 , where C is a constant. 3 dx x y (3) Equation solvable for y: i.e., y = f (x, p), differentiate both sides w.r.t. x we obtain dy ∂f ∂f dp dp = + ≡ F (y, p, ), dx ∂x ∂p dx dx and we solve for p = F (y, p,
dp ), dx
to obtain φ(x, p, C) = 0. Example. Solve the following D.E. y = 2px + p4 x2 . 36
Here we have y = f (x, p) and we therefore differentiate both sides w.r.t. x to obtain dp dp dy = p = 2p + 2x + 4p3 x2 + 2xp4 dx dx dx which when arranged gives (p + 2x and has the solution p + 2x
dp )(1 + 2p3 x) = 0, dx
dp dp dx = 0, −2 = . dx p x
Integrate this to get x=
C . p2
To find y we substitute x in our original D.E. by writing the equation as y − p4 x2 = 2px or (y − p4 x2 )2 = 4p2 x2 , hence, (y − C 2 )2 = 4Cx as a required primitive to our D.E.
3.0.4
Clairaut’s Equation
Is a D.E. of the form y = px + f (p). Its primitive is y = Cx + f (C). Examples. (i) Solve the following D.E. q
y = px + 37
4 + p2 .
Here
q
f (p) =
4 + p2 , so that f (C) =
and hence, y = Cx + f (C) = Cx +
√
√
4 + C 2,
4 + C 2.
(ii) Solve the following D.E. (y − px)2 = (1 + p2 ). We write this as
q
y − px = For
1 + p2 .
q
q
1 + p2 , or y = px +
y − px =
1 + p2 ,
we have the solution y = Cx + f (C) = Cx +
√
1 + C 2,
or (y − Cx)2 = (1 + C 2 ). (iii) Solve the following D.E. y = 3px + 6y 2 p2 . This can be put in the Clairaut’s equation form by multiplying by y 2 , we then obtain y 3 = 3py 2 x + 6y 4 p2 . Now let y 3 = v, so that
dv 2 dv 2 + ( ), dx 3 dx to obtain v = P x + 23 P 2 , which has the solution v=x
we can set P =
dv dx
2 v = Kx + f (K) = Kx + K 2 , K = constant. 3 38
Or y 3 = 3Cx + 6C 2 , where we have put K = 3C. Exercise 3.2 Obtain the primitive of the following D.E. (p =
dy ) dx
(i) y = px − 2p2 , (ii) xp2 − 2yp + 4x = 0,
(iii) x4 p2 − xp − y = 0,
(iv) y 2 p2 + 3px − y = 0, (v) p2 − xp + y = 0, (vi) y = 2px + y 2 p3 , (vii) xp = y +
3.0.5
√
x2 + y 2 .
Ricatti’s Equation
This has the form y 0 + a(x)y 2 + b(x)y + c(x) = 0.
(3.3)
This equation cannot in general be solved by integration. However, it has the property that, given one particular solution y1 , the general solution can be found. If we set y = y1 + z, then y 0 + a(x)y 2 + b(x)y + c(x) = y10 + z 0 + a(x)(y12 + 2y1 z + z 2 ) + b(x)(y1 + z) + c(x), y 0 +a(x)y 2 +b(x)y+c(x) = y10 +a(x)y12 +b(x)y1 +c(x)+[z 0 +a(x)(2y1 z+z 2 )+b(x)z], y 0 + a(x)y 2 + b(x)y + c(x) = z 0 + z[2a(x)y1 + b(x)] + az 2 = 0.
(3.4)
This is Bernoulli equation with n = 2. Example: Find the general solution of Ricatti equation, given that y1 = − x1 is a solution, y 0 + 2y 2 − 39
3 = 0. x2
From equation (3.4) one observes that a = 2, b = 0 and c = − x32 . This can be written as 4 z 0 + (2ay1 + b)z + az 2 = z 0 − z + 2z 2 = 0. x This is Bernoulli equation with n = 2 whose solution is given in sect. 3.0.1
3.0.6
Lagrange’s Equation
This has the form y = xf (y 0 ) + g(y 0 ) , y 0 = p. This is a generalization of Clairaut’s equation, which can be obtained for f (y 0 ) = y 0 , g(y 0 ) = f (y 0 ). We introduced the function p, as before, and consider x as a function of p, i.e., x = x(p). It then follows that dx dx dy 1 dy 1 d(xf (p) + g(p)) 1 dx = = = = [xf 0 (p) + f (p) + g 0 (p)], dp dy dp p dp p dp p dp or
dx f 0 (p) g 0 (p) =x + . dp p − f (p) p − f (p)
This is a linear equation for x = x(p), where x is considered as an independent variable and p the dependent variable. The solution can then be given in the parametric form x = x(p) y = x(p)f (p) + g(p). Example: Solve the following D.E. y = 2xp + 3x + p2 + 6p.
40
This equation can be written as y = x(2p + 3) + p2 + 6p. Comparison with Lagrange equation yields f (p) = 2p + 3 , f 0 (p) = 2 and g(p) = p2 + 6p , g 0 (p) = 2p + 6. Thus, dx f 0 (p) g 0 (p) 2 2p + 6 2 =x + = −x − = −x − 2, dp p − f (p) p − f (p) p+3 p+3 p+3 or
dx 2 =− − 2, dp p+3
which is a first order linear D.E.( where x is the dependent variable and p is the independent variable), i.e., 2 dx + x = −2, dp p + 3 whose solution is (see equation (3.1)) 2 x = − (p + 3) + C(p + 3)−2 , 3 we have thus expressed x in terms of p (x = x(p)) Exercise 3.3 (1) For each of the following D.E., find a solution satisfying the given initial conditions (y 0 =
dy ) dx
(a) y 0 = sin x , y(0) = 1, (b) y 0 = y , y(0) = 1, (c) y 00 = ex , y(1) = 1, y 0 (1) = 0 (2) Find all solutions of the following equations by separation of variables: (i) y 0 = ex+y , (ii) y 0 = 3y, (iii) y 0 = (y − 1)(y − 2), (iv) y 0 = x3 y −2 ,
41
(v) sin x cos y dx + tan y cos x dy = 0, (vi) y 0 = xy 2 + y 2 + xy + y. (3) Solve the following equation after verifying that they are homogeneous: (i) y 0 =
x−y , ; (ii) xy 0 − y = xey/x , (iii) (3x2 y + y 3 )dx + (x3 + 3xy 2 )dy = 0, x+y
(iv) y 0 = xy sin y−x . x (4) Show that the Ricatti’s equation y 0 + a(x)y 2 + b(x)y + c(x) = 0 1
can be transformed into a separable equation by the transformation y = ( ac ) 2 u, if 3
ac > 0 and if (a0 c − ac0 − 2abc)(ac)− 2 = const. Study Ricatti equation y 0 + ay 2 = bxm where a, b are constant, as follows: (a) Apply the transformation u = xy, t = xm+2 . (b) Solve the D.E. tu0 − 12 u + βu2 = γt, β, γ=const. (c) Transform the equation tu0 + αu + βu2 = γt by u =
t ,v a+v
=
1+α , γ
and
u = a + vt , a = − αβ (d) Conclude that y 0 + ay 2 = bxm can be solved in a closed form for m = 4n − 2n+1 , where n is an integer. (e) Solve the equations y 0 + y 2 = x−4 ,
42
xy 0 + 3y + y 2 = x2 .
Chapter 4 Higher Order Linear Differential Equations A linear D.E. of order n has the form dn y dn−1 y dn−2 y + a + a + .... + an y = Q(x) 1 2 dxn dxn−1 dxn−2 where a1 , a2 , a3 , ... are functions of x or constants, and all terms of first degree (linear). When Q = 0 the D.E. is called homogeneous, and when Q 6= 0 it is called non-homogeneous. Example. d2 y dy − 2x + y = 3x + 1, 2 dx dx is a non homogeneous linear D.E.(second order) and d2 y dy + + x2 y = 0, dx2 dx is a homogeneous linear D.E. (second order). If y1 , y2 , y3 are solutions to the D.E. then c1 y1 + c2 y2 + c3 y3 , where c1 , c2 , c3 are arbitrary constants, is also a solution. The set of solutions y1 , y2 , y3 are called linearly independent if c1 y1 + c2 y2 + c3 y3 = 0, 43
implies that c1 = c2 = c3 = 0, as the only solution. Example. If y1 = ex ,
y2 = e−x ,
show that the two solutions are linearly independent. c1 y1 + c2 y2 = 0, and by differentiating, we get c1
dy1 dy2 + c2 = 0. dx dx
Then, c1 ex + c2 e−x = 0, and c1 ex − c2 e−x = 0. Solving the two equations, we get c1 = c2 = 0. Hence, the two solutions are linearly independent. A necessary and sufficient condition for a set of solutions be linearly independent is that
¯ ¯ y1 ¯ ¯ 0 ¯ y1 W = ¯¯ y 00 ¯ 1 ¯ .. ¯ .
y2 y20 y200 .. .
¯
. . . ¯¯ . . . ¯¯ ¯ . . . ¯¯ 6= 0 .. ¯¯ .
y3 y30 y300 .. .
(W is called the Wronskian), where y0 =
dy , dx
y 00 =
d2 y , ... dx2
Example. Show that the two solutions e−x , 44
e2x ,
are linearly independent. ¯ ¯y W = ¯¯ 10 y 1
¯
¯
y2 ¯¯ ¯¯ e−x = y20 ¯ ¯ −e−x
¯
e2x ¯¯ = 3ex 6= 0, 2e2x ¯
hence, the two solutions are linearly independent. Exercise. Show that y = eax is a solution of the D.E. d2 y dy − − 2y = 0, dx2 dx for some values of a and find these values. Exercise 4.1 (1) Compute the Wronskian of the following pairs of functions, and show which of the pairs are linearly independent. sin ax, cos bx; sin2 x, 1−cos 2x; x, x ln x; eax , xeax ; eax , ebx ; eax sin bx, eax cos bx where a, b are constants. (2) Show that if y1 and y2 are solution of the D.E y 00 + p(x)y 0 + q(x)y = 0 then their Wronskian (W ) satisfies the D.E. W 0 + p(x)W = 0, where W 0 = dW . dx
4.1
Homogeneous D.Es. with constant coefficients
This can be written in the form dn y dn−1 y dn−2 y + a + a + .... + an y = 0, 1 2 dxn dxn−1 dxn−2
(4.1)
where a1 , a2 , a3 , ..., an are constant coefficients. Define the operator D as D = then the above equation becomes Dn y + a1 Dn−1 y + a2 Dn−2 y + ... + an y = 0, 45
d , dx
which can be written as L(D)y = 0, where L(D) = Dn + a1 Dn−1 + a2 Dn−2 + ... + an , which is function of D. The equation L(D)y = 0, which has the general solution (D − m1 )(D − m2 )(D − m3 ).....(D − mn )y = 0, is called the characteristic equation and the roots m1 , m2 , m3 , ... are called the characteristic roots. The primitive y obtained from this equation is called the complementary solution. Alternatively, one can assume the solution to have the form y = emx and substitute this in equation (4.1) to obtain an equation for m. Solution of m gives the roots of the equation. The general solution is the sum of these solutions (emx ). The operator D has the following properties: (i) D[cy] = cD[y] (ii) D[y1 + y2 ] = D[y1 ] + D[y2 ] such an operator which satisfies the above properties is called a linear operator. All other operators are nonlinear. Example. Write the D.E. in the operator form and find the characteristic roots d2 y dy d3 y − − 4 + 4y = 0, 3 2 dx dx dx becomes (D3 − D2 − 4D + 4)y = 0, 46
that can be put in the form (D − 1)(D − 2)(D + 2)y = 0, which has the characteristic roots: m1 = 1, m2 = 2, m3 = −2.
4.1.1
The complementary solution of a D.E.
(a) If the roots are real and different, i.e., m1 6= m2 6= m3 .... then the complementary solution is y = c1 em1 x + c2 em2 x + c3 em3 x + ... (b) If the roots are the same, i.e., m1 = m2 6= m3 .... the complementary solution is y = c1 em1 x + c2 xem1 x + c3 em3 x + ... (c) If the roots are complex, i.e., m1 , m2 = a ± ib, where i =
√
−1, the comple-
mentary solution is y = eax (c1 sin bx + c2 cos bx). Examples. Find the complementary solution of the following D.Es. (i) d3 y d2 y dy − 2 − 4 + 8y = 0. 3 2 dx dx dx This can be written as (D3 − 2D2 − 4D + 8)y = (D − 2)(D − 2)(D + 2)y = 0, hence, m1 = 2, m2 = 2, m3 = −2, thus two roots are equal and, therefore, the complementary solution is y = c1 e2x + c2 xe2x + c3 e−2x . 47
(ii)
d2 y dy − 5 + 4y = 0. 2 dx dx
This can be written as (D2 − 5D + 4)y = (D − 4)(D − 1)y = 0, hence, the roots are m1 = 4, m2 = 1 (the roots are different) and the complementary solution is y = c1 e4x + c2 ex . (iii)
d3 y d2 y dy − − 12 = 0, dx3 dx2 dx
which can be written as (D3 − D2 − 12D)y = D(D2 − D − 12)y = D(D − 4)((D − 3)y = 0, hence, the roots are m1 = 0, m2 = 4, m3 = 3, and the complementary solution y = c1 + c2 e4x + c3 e3x . (iv)
d2 y dy − 2 + 10y = 0, 2 dx dx
can be put in the form (D2 − 2D + 10)y = 0, the roots are m1 , m2 = 1 ± 3i, and hence, the complementary solution y = ex (c1 sin 3x + c2 cos 3x). (v) (
d2 y dy − 2 + 5)2 y = 0, 2 dx dx 48
this, in operator form, becomes (D2 − 2D + 5)2 y = (D2 − 2D + 5)(D2 − 2D + 5)y = 0. Hence, the roots are m1 , m2 = 1 ± 2i and m3 , m4 = 1 ± 2i, which are the same (repeated), therefore the complementary solution is y = ex (c1 sin 3x + c2 cos 3x) + xex (c3 sin 3x + c4 cos 3x). Exercise 4.2 Find the complementary solutions of the following D.Es. (1) (i)
d3 y dx3
d y − 2 dx 2 −
2
(ii)
d2 y dx2
dy − 4 dx + 4y = 0,
dy dx
+ y = 0,
(iii)
d2 y dx2
+
(iv)
d3 y dx3
d y dy − 2 dx 2 − 4 dx + 8y = 0,
(v)
d2 y dx2
dy dx
− 2y = 0,
2
dy + 2 dx + 2y = 0.
(2) Let D[y](x) = y 00 − 3xy 0 + 3y, compute D[ex ] , D[x2 ] , D[x] , D[x2 + 3x] Let D[y](x) = y 00 + p(x)y 0 + q(x)y and suppose that D(x2 ) = x + 1 and D(x) = x + 2, show that: y(x) = x − 2x2 is a solution of the above D.E. √ (3) Show that y1 = x and y2 =
1 x
are solutions of the D.E.
2x2 y 00 + 3xy 0 − y = 0. (4) Show that the operator D defined by D[y] =
Z b a
s2 y(s)ds
is linear, that is D[cy] = cD[y] and D[y1 + y2 ] = D[y1 ] + D[y2 ]. 49
4.2
Non homogeneous linear D.Es.
4.2.1
Operator Method
This has the form L(D)y = Q(x), which can be written as y= or y=
1 Q(x) L(D)
1 Q(x), (D − m1 )(D − m2 )(D − m3 ).....(D − mn )
or
1 1 1 1 ... Q(x). D − m1 D − m2 D − m3 D − mn To find the solution y, we employ the following method: y=
(4.2)
Set
Q du , or (D − mn )u = Q, − mn u = Q. D − mn dx One can then solve this linear first order D.E. to find u (the solution is u = u=
R
emn x Qe−mn x dx, see equation (3.1)) and then write w=
u D − mn−1
or
dw − mn−1 w = u dx this latter equation can be solved to obtain w. We continue like this until one reaches the final term in equation (4.2), and finally we solve for y. The general solution is the sum of the complementary solution (hereafter denoted by yc ) that obtained from L(D)y = 0 and the above solution, which is called the particular solution (hereafter denoted by yp ) (for the non-homogeneous part), thus y = yc + yp . Examples. (i) Find the general solution of the following D.E. (D2 − 3D + 2)y = ex . 50
The complementary solution (yc ) is obtained from the homogeneous part, i.e., L(D)y = 0 and the particular solution is due to the non-homogeneous part. Thus (D − 2)(D − 1)y = 0, gives yc = c1 ex + c2 e2x . To find the particular solution (yp ) we proceeds as follows: Write yp =
1 1 ex , D−1D−2
and set u=
ex , D−2
yp =
u . D−1
hence, Thus one obtains
du − 2u = ex , dx
which has the solution Z
u = e2x Now yp =
ex e−2x dx = −ex . u −ex = , D−1 D−1
or
−ex , D−1 which has the solution yp =
Dyp − yp = −ex ,
dyp − yp = −ex , dx
Z
yp = e x
(−ex e−x )dx = −xex .
Thus the particular solution is yp = −xex and hence, the general solution is y = yc + yp = c1 ex + c2 e2x − xex .
51
(ii) Find the general solution of the following D.E. (D3 + 3D2 − 4)y = xe−2x . This can be written as (D − 1)(D + 2)(D + 2)y = xe−2x . The complementary solution yc = c1 ex + c2 e−2x + c3 xe−2x . To find the particular solution, we write yp =
1 1 xe−2x , D−1D+2D+2
and set u= to get
xe−2x , D+2
du + 2u = xe−2x , dx
which has the solution Z
1 xe−2x e2x dx = x2 e−2x . 2
u = e−2x Now yp =
1 1 1 1 1 u= ( x2 e−2x ), D−1D+2 D−1D+2 2
set w= so that
1 1 2 −2x xe , D+22
1 (D + 2)w = x2 e−2x , 2
which can be solved to get Z
w = e−2x
1 1 2 −2x 2x x e e dx = x3 e−2x , 2 6 52
and finally, we get yp = or
1 w, D−1
so that yp =
1 3 −2x xe 6
D−1
.
1 Dyp − yp = x3 e−2x , 6
which has the solution Z
yp = ex
1 3 −2x −x 1 2 2 x e e dx = − (x3 + x2 + x + )e−2x . 6 18 3 9
Hence,
1 3 2 2 (x + x2 + x + )e−2x . 18 3 9 and the general solution is given by yp = −
y = yc + yp = c1 ex + c2 e2x −
1 3 2 2 (x + x2 + x + )e−2x . 18 3 9
(iii) Find the general solution of the following D.E. (D2 − 3D + 2)y = e5x . This can be written as (D − 1)(D − 2)y = e5x . The complementary solution is y = c1 ex + c2 e2x . To find the particular solution, we write y=
1 1 e5x , D−1D−2
and set u=
e5x , D−2 53
so that
du − 2u = e5x , dx
which has the solution Z
u = e2x and hence, y=
1 e5x e−2x dx = e5x , 3
1 e5x 1 1 u= ⇒ Dy − y = e5x , D−1 3D−1 3
which yields
Z
yp = ex
1 5x −x 1 e e dx = e5x . 3 12
Thus, the general solution is y = c1 ex + c2 e2x +
1 5x e . 12
Exercise 4.3 (i) Show that the particular solution of the D.E. (D − a)(D − b)y = Q, is given by
Z
y = eax
Z
e(b−a)x [ebx
Qe−bx dx]dx.
(ii) Find the general solution of the D.E. (D2 + 5D + 4)y = 3 − 2x. (iii) Find the general solution of the D.E. (D2 − 1)y = 4xex . (iv) Find the general solution of the D.E. (D3 − 4D)y = x. (v) Find the general solution of the D.E. (D3 − 5D2 + 8D − 4)y = e2x .
4.2.2
The properties of L(D)
If L(D) is a differential operator with constant coefficients, one can write L(D) =
n X r=0
54
ar Dr
If y1 , y2 are solutions then L(D)(c1 y1 + c2 y2 ) = c1 L(D)y1 + c2 L(D)y2 . This follows since L(D)(c1 y1 + c2 y2 ) =
n X
ar Dr (c1 y1 + c2 y2 )
r=0
= c1
n X
ar Dr y1 + c2
r=0
n X
ar Dr y2 = c1 L(D)y1 + c2 L(D)y2 .
r=0
Theorem I: L(D)eax = L(a)eax , (a = const.). Theorem II: L(D)[eax f (x)] = eax L(D + a)f (x). Theorem III: L(D2 ) sin ax = L(−a2 ) sin ax,
L(D2 ) cos ax = L(−a2 ) cos ax.
We define the inverse operator L−1 (D) or 1/L(D) by means of (1) L(D)[L−1 (D)]f (x) = f (x). Thus if yp = L−1 (D)f (x), then L(D)yp = f (x) and yp is a particular integral. If f (x) is a polynomial one can expand L(D) in binomial expansion. Thus, 1 1 1 =− [ D−a a 1−
D a
1 D D D ] = − [1 + + ( )2 + ( )3 + ...]. a a a a
Example: Solve the D.E. (D − 1)(D − 3)y = x2 + 2x3 . 55
In this case we have the particular solution y=
x2 + 2x3 1 1 1 = ( − )(x2 + 2x3 ). (D − 1)(D − 3) 2 D−3 D−1
Hence, 1 1 D D D y = [− (1 + + ( )2 + ( )3 + ... + (1 + D + D2 + D3 + ...)][x2 + 2x3 ] 2 3 3 3 3 1 4 13 40 186 20 2 = ( + D + D2 + D3 + ...)[x2 + 2x3 ] = + x + 3x2 + x3 . 3 9 27 81 27 3 3 3 We stopped at D since all higher derivatives give zero contribution. (2)
1 sin ax sin ax = . 2 L(D ) L(−a2 )
Solve (D − 2)(D − 3)y = sin 3x. The particular solution is yp =
(D2
1 1 −1 sin 3x = sin 3x = sin 3x, 2 − 5D + 6) −3 − 5D + 6 3 + 5D =−
= (3)
3 − 5D 3 − 5D sin 3x = − sin 3x, 2 9 − 25D 9 − 25.(−32 )
1 1 (3 − 5D) sin 3x = (3 sin 3x − 15 cos 3x). 234 234 1 ax 1 ax e = e , L(a) 6= 0. L(D) L(a)
Solve the following D.E. (D2 − 5D + 6)y = e5x . yp =
D2
1 1 5x e5x = e − 5D + 6 L(5) 56
yp =
1 1 5x 5x e = e , 52 − 5.5 + 6 6
(where L(D) = D2 − 5D + 6 and L(5) = 52 − 5.5 + 6 = 6). (4) (i)
1 1 [eax f (x)] = eax f (x), L(D) L(D + a)
(ii)
1 xm eax ax e = . (D − a)m L(D) m!L(a)
(ii) Solve the D.E. (D2 − 5D + 6)y = e4x x2 . yp =
D2
1 1 1 e4x x2 = e4x x2 = e4x 2 x2 , 2 − 5D + 6 (D + 4) − 5(D + 4) + 6 (D + 3D + 2)
1 1 1 4x 1 2 1 2 2 yp = e4x x = e [1 − (D + 3D) + (D + 3D)2 + ..]x2 , 1 2 [1 + 2 (D2 + 3D)] 2 2 4 1 3 7 yp = e4x (1 − D + D2 + ..)x2 . 2 2 4 Hence,
1 7 yp = e4x (x2 − 3x + ). 2 2
(iii) Solve the D.E. (D − 1)(D − 3)y = e3x . yp =
1 1 e3x e3x = [ ] (D − 1)(D − 3) D−1 D−3 1 1 e3x [ .1] D−1 D+3−3 1 3x 1 1 yp = e [ .1] = xe3x , 3−1 D 2 yp =
where we have used and the fact that
1 e3x D−3
1 .1 D
=
1 [e3x .1] D−3
(we have put f (x) = 1, in (4) above)
= x. We use this trick when L(a) = 0.
More generally, we can evaluate 1 1 f (x) = eλx (e−λx f (x)) D−λ D−λ 57
eλx
Z 1 1 (e−λx f (x)) = eλx (e−λx f (x)) = eλx e−λx f (x)dx. D−λ+λ D
Hence,
Z 1 λx f (x) = e e−λx f (x)dx. D−λ
(4.3)
(iv) Solve the D.E. (D2 − 3D + 2)y = (D − 2)(D − 1)y = e3x . The particular solution is yp =
1 3x 1 3x e = e , where, L(3) = 32 − 3.3 + 2. L(3) 2
(v) Solve the D.E. (D − 2)(D − 3)y = e4x sin 3x. sin 3x e4x sin 3x = e4x (D − 2)(D − 3) (D + 2)(D + 1) sin 3x sin 3x sin 3x yp = e4x 2 = e4x = e4x 2 (D + 3D + 2) (−3 + 3D + 2) (3D − 7)
yc = c1 e2x + c2 e3x ,
yp = e4x
yp =
3D + 7 3D + 7 1 4x sin 3x = e sin 3x = (9 cos 3x+7 sin 3x)e4x . (9D2 − 49) 9(−32 ) − 49] −130
(vi) Solve the D.E. y 00 + 3y 0 − 4y = sin 2x. The particular solution is yp =
1 1 1 1 sin 2x = ( − ) sin 2x, (D − 1)(D + 4) 5 D−1 D+4
1 1 1 1 1 x Z −x 1 −4x Z 4x yp = sin 2x− sin 2x = e e sin 2x dx− e e sin 2x dx. 5D−1 5D+4 5 5 Using equation (4.3). (5) (i)
D2
1 x cos ax sin ax = − , 2 +a 2a 58
D2
1 x sin ax cos ax = , 2 +a 2a
1 sin bx 1 cos bx sin bx = 2 , cos bx = 2 , a 6= b, 2 2 2 2 +a a −b D +a a − b2 Z Z Z 1 1 1 Z (iii) f (x) = f (x)dx, 2 f (x) = f (x)dx = [ f (x)dx]dx. D D D (vii) Solve the following D.Es. (ii)
D2
(a) (D2 + 4)y = sin 2x, (b) (D2 − 2D + 1)y = 2xex . 1 sin 2x D2 +22 x 1 2e D2 x = 2ex D1
(a) yp =
= − 41 x cos 2x, (b) yp = R x
R
yp = xdx = ex D1 x2 = e One can accordingly solve example (vi) as yp =
1 2xex = D2 −2D+1 x2 dx = 31 ex x3 .
1 2ex (D+1)2 −2(D+1)+1 x,
(D + 1)(D − 4) (D + 1)(D − 4) sin 2x = sin 2x, 2 2 2 (D − 1)(D − 4 ) (−22 − 1)(−22 − 42 )
yp =
1 1 (D2 − 3D − 4) sin 2x = − (4 sin 2x + 3 cos 2x), 100 50
Exercise 4.4 (1) Prove Theorem I, II, III stated before. (2) Find the general solution of the following D.Es. (i) y 00 + 36y = 2 sin 3x, (ii) y 00 − 4y 0 + 4y = 3 sin 9x , (iii) y 00 + 4y 0 + 13y = 2x + 7e3x , (iv) y 00 + 5y 0 + 6y = 2 cosh x. (3) Show that
4.3
1 eax (D−a)m
=
xm eax m!
and
1 eax (D−a)m L(D)
=
xm eax . m!L(a)
Solution of D.Es. by variation of parameters method
The D.E. can be put in the general form L(D)y = Q(x). The complementary solution has the form yc = c1 y1 + c2 y2 + c3 y3 + ...
59
We now allow the constant (c’s) to vary and write instead, y = L1 (x)y1 + L2 (x)y2 + L3 (x)y3 + ... as our particular solution. The problem is to find the parameters L1 , L2 , L3 , ... as dy d2 y d3 y 000 functions of x from the following equations:(y 0 = dx , y 00 = dx = dx 2,y 3) y 0 = L01 y1 + L02 y2 + L03 y3 + (L1 y10 + L2 y20 + L3 y30 ), and set L01 y1 + L02 y2 + L03 y3 = 0,
(4.4)
y 00 = L01 y10 + L02 y20 + L03 y30 + (L1 y100 + L2 y200 + L3 y300 ), and set L01 y10 + L02 y20 + L03 y30 = 0,
(4.5)
y 000 = L01 y100 + L02 y200 + L03 y300 + (L1 y1000 + L2 y2000 + L3 y3000 ), and set L01 y100 + L02 y200 + L03 y300 = Q,
(4.6)
so that, if we write L(D) = D3 + P1 D2 + P2 D + P3 , L(D)y = Q, and substitute for D3 y, D2 y, Dy, we obtain L1 (y1000 + P1 y100 + P2 y10 + P3 y1 ) + L2 (y2000 + P1 y200 + P2 y20 + P3 y2 ) +L3 (y3000 + P1 y300 + P2 y30 + P3 y3 ) + Q = 0, and since y1 , y2 , y3 are solutions of the equation L(D)y = 0, all terms in the brackets are zero. Examples. Find the general solution of the following D.Es. by the method of variation of 60
parameters (i) (D2 − 2D)y = ex sin x. The complementary solution is yc = c1 + c2 e2x , so that we write for the particular solution the form y = L1 + L2 e2x . Apply the conditions (4.4)-(4.6) above, we obtain L01 + L02 e2x = 0, Hence,
2L02 e2x = Q = ex sin x.
1 L02 = e−x sin x, 2
which can be integrated to yield L2 =
1 Z −x 1 e sin x dx = − e−x (sin x + cos x), 2 4
and hence,
1 L01 = −L02 e2x = − ex sin x, 2 which can be integrated to get 1Z x 1 L1 = − e sin x dx = − ex (sin x − cos x). 2 4 Thus
1 yp = L1 + L2 e2x = − ex sin x, 2 and finally the general solution is 1 y = c1 + c2 e2x − ex sin x. 2 61
(ii) (D3 + D)y = csc x. This can be written as D(D2 + 1)y = csc x, which has the complementary solution yc = c1 + c2 cos x + c3 sin x, so that the particular solution is of the form y = L1 + L2 cos x + L3 sin x. To determine L1 , L2 , L3 we apply the conditions (4.4)-(4.6) above. Hence, (i) L01 + L02 cos x + L03 sin x = 0, (ii) − L02 sin x + L03 cos x = 0, (iii) − L02 cos x − L03 sin x = Q = csc x. Adding (i) and (iii), we get: L01 = csc x and hence, L01 = − ln(csc x + cot x), solving (ii) and (iii) we obtain: L03 = −1, hence, L3 = −x and L02 = − cot x , L2 = − ln(sin x). Thus yp = L1 + L2 cos x + L3 sin x = − ln(csc x + cot x) − ln(sin x) cos x − x sin x, and finally, the general solution is y = c1 + c2 cos x + c3 sin x − ln(csc x + cot x) − ln(sin x) cos x − x sin x.
4.4
Solution of D.Es. by undetermined coefficients method
If L(D)y = x3 , we can write the particular solution as yp = Ax3 + Bx2 + Cx + E, 62
where A, B, C and E are undetermined coefficients. If L(D)y = ex + e3x , then the particular solution of the D.E. takes the form yp = Aex + Be3x . Special Cases: (1) If a term of Q is also a term of yc , say u, which has a s-fold root, then we introduce a term of xs u plus terms arising from it by differentiation Example. (D − 2)2 (D + 3)y = e2x + x2 , then yc = c1 e2x + c2 xe2x + c3 e−3x , then the term e2x appears in Q and in yc in this case yp has a term with multiplicity of 2, hence, yp should have a term x2 e2x and all terms arising from it by differentiation. Thus, yp = Ax2 e2x + Bxe2x + Ce2x + Ex2 + F x + G. (2) If a term of Q is a xr u and u is a term of yc , if u corresponds to s-fold root, yp must contain a tern xr+s u plus all terms arising from it by differentiation. Examples. (1) Solve the following D.E. (D − 2)3 (D + 3)y = x2 e2x + x2 , the root m = 2 is three fold (three equal roots), i.e., s = 3. The complementary solution is yc = c1 e2x + c2 xe2x + c3 x2 e2x + c4 e−3x , and we see that the term x2 e2x (in Q) is also in yc , hence, yp must contain a term x2+3 e2x , and all terms arising from it by differentiation. Thus yp = Ax5 e2x + Be2x x4 + Cx3 e2x + Gx2 + Hx + J. 63
(the terms Ex2 e2x + F xe2x + M e2x are all already in the RHS). (2) Solve the D.E. (D2 − 2D + 3)y = x3 + sin x. √ This has the roots m1 , m2 = 1 ± 2 i, so that the complementary solution is √ √ yc = ex (c1 sin 2x + c2 cos 2x). We write the particular solution as yp = Ax3 + Bx2 + Cx + E + F sin x + G cos x. Substitute this in our differential equation to determine the constant A, B, C, E, F, G. Now Dyp = 3Ax2 +2Bx+C +F cos x−G sin x, D2 yp = 6Ax+2B−F sin x−G cos x. Hence, (6Ax + 2B − F sin x − G cos x) − 2(3Ax2 + 2bx + C + F cos x − G sin x) +3(Ax3 + Bx2 + Cx + E + F sin x + G cos x) = x3 + sin x. Equating the coefficient of the different expressions in both sides yields 1 2 2 16 1 1 A = ,B = ,C = − ,E = − ,F = ,G = . 3 3 9 27 4 4 (3) Solve the D.E. (D2 + 4)y = x2 sin 2x. The roots for (D2 + 4)y = 0 are m1 , m2 = ±2 i, yc = c1 sin 2x + c2 cos 2x, but the term sin 2x in Q is also in yc with multiplicity of 1 (1-fold). Hence, yp must contain a term x2+1 sin 2x = x3 sin 2x. Therefore yp = Ax3 sin 2x + Bx2 sin 2x + Cx sin 2x + Ex3 cos 2x + F x2 cos 2x + Gx cos 2x. 64
Note that sin 2x + cos 2x are not included in yp since they are already in yc . Substitute y = yp in the original D.E above and equate the coefficients of different expressions to get A, B, C, E, F, G to complete the solution (A = 0, B =
1 ,C 16
=
1 1 0, E = − 12 , F = 0, G = − 32 ).
Exercise 4.5 (1) Solve the following D.Es by the method of undetermined coefficients (i) (D2 + 2)y = ex + 2, (ii) (D2 + 2D + 2)y = sin x + x2 , (iii) (D2 − 1)y = ex sin 2x, (iv) (D2 − 5D + 6)y = x2 e4x + 9e4x . (2) Solve the following D.Es by the method of variation of parameters (i) (D2 + 1)y = cos x, (ii) (D3 + 4D2 + 3D)y = x2 , (iii) (D2 + 4)y = sin 2x, (iv) y 00 + y = tan x, (v) (D2 − 6D + 9)y =
4.5
e3x . x2
Euler’s Equation
The homogeneous equation has the form x2
d2 y dy + ax + by = 0, 2 dx dx
(4.7)
where a, b are constants. This can be changed to an equation with constant coefficients upon the transformation x = et so that, t = ln x. Now dy dy dt 1 dy = = , dx dt dx x dt hence, we see that and
dy dy = e−t , dx dt d d = e−t , dx dt 2 d dy d2 y dy −t d −t dy −t −t d y = ( ) = e (e ) = e (e − e−t ), 2 2 dx dx dx dt dt dt dt
65
or e−2t
2 d2 y dy −2t dy −2t d y − e = e ( − ). 2 2 dt dt dt dt
Writing y˙ =
dy , dt
y¨ =
one yields
d2 y , dt2
d2 y 1 = (¨ y − y), ˙ dx2 x2
dy 1 = e−t y˙ = y, ˙ dx x and hence, equation (4.7) becomes
(¨ y − y) ˙ + ay˙ + by = 0, or y¨ + (a − 1)y˙ + by = 0, which has the solution y(t) = c1 em1 t + c2 em2 t , and finally substitute t = ln x in this equation. Examples. (i) Solve the Euler’s equation x2
d2 y dy + 5x + 5y = 0. 2 dx dx
Letting x = et one obtains y¨ + (a − 1)y˙ + by = 0, which becomes (a = 5, b = 5) y¨ + (5 − 1)y˙ + 5y = y¨ + 4y˙ + 5y = 0, which has the solution (in terms of t the roots are m1 , m2 = −2 ± i) y = e−2t (c1 sin t + c2 cos t), 66
or
1 (c1 sin(ln x) + c2 cos(ln x)), since t = ln x. x2 (ii) Solve the non homogeneous D.E. y=
d2 y dy − x + 2y = x ln x. 2 dx dx t Let x = e , the equation becomes x2
y¨ + (a − 1)y˙ + by = x ln x, or y¨ − 2y˙ + 2y = x ln x = tet , since a = −1, b = 2 or y¨ − 2y˙ + 2y = tet which can be solved by one of the previous methods to give y(x) = x(c1 cos(ln x) + c2 sin(ln x)) + x ln x. (iii) Solve the following D.E. d2 y dy + 3x + 2y = x2 + 2x. 2 dx dx Putting x = et , we have (in terms of t) x2
y¨ + 2y˙ + 2y = e2t + 2et , whose complementary solution yc = e−t (A cos t + B sin t) and the particular solu1 2t tion is yp = 10 e + 25 et . Hence, in terms of x, we have y(x) =
1 1 2 (A cos ln x + B sin ln x) + x2 + x. x 10 5
Exercise 4.6 (1) Solve the following differential D.Es. 2
d y dy (i) x2 dx 2 − 5x dx + 9y = 0, 2
2
d y (ii) x2 dx 2 − 2y = 0, 2
d y dy dy 2 2d y (iii) x2 dx 2 + 3x dx − 3y = 5x , (iv) x dx2 + x dx + y = x(6 − ln x), 2
d y (v) x2 dx 2 − 2y = ln x ,
2
d y dy 2 (vi) x2 dx 2 + 5x dx + 13y = ln x + x .
67
4.5.1 Put p =
Second order D.E. of the form dy dx
d2 y dx2
= f (y)
so that d2 y dp dy dp = =p . 2 dx dy dx dy
Thus p
dp = f (y) , dy
or
Z 1 2 p = A + f (y)dy, 2
hence we can find y(x). Example: Solve
d2 y = y 2 given that y 0 (y = 0) = 0, y(x = 0) = ∞. dx2
Writing p =
dy dx
gives p
dp 1 2 1 3 = y2 , p = y + C. dy 2 3
√ But p(y = 0) = 0, y(x = 0) = ∞ gives C = 0 and hence x = ∓ 6y −1/2 .
4.5.2
Second order D.E. of the form
The assumption p =
dy dx
d2 y dx2
dy = f (y, dx )
dp gives p dy = f (y, p) which could be integrable.
Example: Solve yy 00 = y 02 − y 03 . Put p =
dy dx
dp so that y 00 = p dy . Hence
yp
dp dp = p2 − p3 , p(y − p + p2 ) = 0, dy dy
dp so that either p = 0 which implies that y = const. or y dy = p − p2 , which implies dy y
=
dp p
+
dp , 1−p
p which upon integration gives y = C 1−p or p =
grating this one obtains C ln y + y = x − B, C, B = const.
68
dy dx
=
y . C+y
Inte-
Exercise 4.7 (1) Find the general solution of yy 00 + 2y 02 = 0. (2) Find the general solution of the following D.Es. d2 y +y = dx2 d2 y dy (iii) dx2 + dx d2 y dy (v) dx 2 + 3 dx
(i)
2
d y dy 4 cot x, (ii) x dx 2 + 6 dx = x , 2
d y + 4y = ex , (iv) dx 2 − y = x + 1, dy d2 y + 2y = sin x, (vi) dx 2 + dx − 2y = 2x,
and find the particular solution if (vii)
d2 y dx2
dy dx
= 0, y = 1 at the point x = 0,
+ y = 2 cos x,
dy and find the particular solution if y = 0, dx = 1 at x = 0.
(3) Solve the Euler’s equation x2 y 00 − 2nxy 0 + n(n + 1)y = 0. (4) Find the transformation y = v(x)g(x) such that the equation y 00 + p(x)y 0 + q(x)y = 0, would become v 00 + r(x)v = 0, and find r(x) in terms of p and q. (5) Given y 00 + P (x)y 0 + Q(x)y = R(x). (a) Prove that if Q − 21 P 0 − 14 P 2 is a constant, the given equation may be solved R
by putting y = u(x) exp(− 12 P (x)dx). Hence, solve the equations (i) y 00 − 2y 0 cot x + 2y cot2 x = sin 2x, (ii) x2 y 00 − 2x2 y 0 + [x2 − n(n + 1)]y = 0. (b) Letting t(x) be the new independent variable, show that the equation reduces to
d2 y 00 0 dy + (t + P t ) + Qy = R, dt2 dt dt 0 02 where t = dx , etc. Now choose t so that t = kQ, where k is a constant. t02
Prove that if (t00 + P t0 )/t02 is a constant, then the equation reduces to one of the 69
constant coefficients. (c) Use the method to solve y 00 + (2 cos x + tan x)y 0 + y cos2 x = cos4 x. (6) Let y1 and y2 be solutions of Bessel’s equation x2 y 00 + xy 0 + (x2 − n2 )y = 0, with y1 (1) = 1, y10 (1) = 0, y2 (1) = 0 and y20 (1) = 1, compute the Wronskian for y1 and y2 . (7) Show that y = xr is a solution of the D.E. x2 y 00 + αxy 0 + βy = 0, provided that r2 + (α − 1)r + β = 0 and hence, find the solutions of the following D.Es. (i) x2 y 00 + 5xy 0 − 5y = 0, (ii) x2 y 00 − xy 0 − 2y = 0; y(1) = 0, y 0 (1) = 1. (8) Given the equation xy 00 − (1 + 3x)y 0 + 3y = 0, has a solution of the form ecx , for some constant c, find the general solution. (9) Find the general solutions of the following D.Es. (i) x2 y 00 + 3xy 0 + y = 0, (ii) x2 y 00 − xy 0 + y = 0. (10) Show that if y1 is a solution of y 00 + p(x)y 0 + q(x)y = 0, R
e
−
R
(4.8)
p(x)dx
then y2 = y1 u(x)dx, where u = y2 (x) , is also a solution. (Hint use the 1 transformation y2 = y1 v(x).) This method is called the reduction of order, since the substitution y2 = y1 v(x) reduces the problem of solving the second order 70
equation to that of solving a first order equation. Examples: (1) Find the general solution of the following D.E. (1 − x2 )y 00 + 2xy 0 − 2y = 0, y(0) = 3, y 0 (0) = −4. Clearly, y1 = x is one solution of the D.E. R To find the second solution we use the R
e
above formula y2 = y1 u(x)dx, u =
−
in the standard form (equation (4.8)) y 00 + Thus, p(x) =
2x 1−x2
. First of all, we write the equation
2x 0 2 y − y = 0. 2 1−x 1 − x2
and hence, u =
e
−
Z
y2 = y1
p(x)dx
y12 (x)
R
Z
u dx = x
2x dx 1−x2 x2
=
1−x2 , x2
and
1 − x2 dx = −(1 + x2 ), 2 x
is the second solution. Therefore, the general solution is y = c1 x−c2 (1+x2 ). One can determine the coefficients c1 , c2 according to the boundary conditions stated above. Thus, 3 = y(0) = −c2 and −4 = y(0 0) = c1 . Hence, y = −4x + 3(1 + x2 ). (2) Find the solution of the D.E. x2 y 00 − 7xy 0 + 16y = 0. Clearly, the substitution xk , k = const., is a solution of the D.E. for k = 4, hence the second solution is Z
y2 = y1
Z
udx = x
4
R
R
Z Z 7 Z e x dx x dx e− p(x)dx 4 4 4 dx = x dx = x dx = x = x4 ln x, 2 8 8 y1 (x) x x x
hence, y2 = x4 ln x.
71
7
and the general solution is y = c1 y1 + c2 y2 = c1 x4 + c2 x4 ln x. (3) Find the general solution of the D.E. (x2 + 1)y 00 − 2y = 0. Clearly, y1 = x2 + 1 is a solution of the D.E. and hence one can obtain the second solution. Now p(x) = 0 so that Z
y2 = y1
Z
udx = y1 Z
y2 = (x2 + 1)
R
Z e0 e− p(x)dx 2 dx = (x + 1) dx y12 (x) (x2 + 1)2
1 x 1 2 dx = (x + 1)[ + tan−1 x], (x2 + 1)2 2(x2 + 1) 2
or
1 1 y2 = x + (x2 + 1) tan−1 x. 2 2
Exercise 4.8 (1) Verify that y1 = x(x − 1)−2 is a solution of the D.E. x(x − 1)y 00 + 3xy 0 + y = 0, and find the second solution. √ (2) Using the substitution t = x, show that the general solution of the D.E. √ √ 4xy 00 + 2(1 − x)y 0 − 6y = e−2 x , is √ √ √ 1√ y = A exp(3 x) + B exp(−2 x) − x exp(−2 x). 5 (3) Find the second solution of the following D.Es. with the given one solutions and hence write down the general solution: (a) y 00 − xy 0 + y = 0, y1 = x; (b) y 00 −
6 y x2
= 0, y1 =
1 . x2 0
(c) (1 − x2 )y 00 − 2xy 0 + 2y = 0 , y1 = x; (d) x2 y 00 + 4xy − 10y = 0, y1 = x2 . (4) Given that the D.E. 4x(1 − x)y 00 + 2(3 − 4x)y 0 − y = 0, 72
has a solution of the form xk , determine k and hence solve the equation completely. (5) Find the complete solution of the D.E. x2 y 00 − 2xy 0 + 2y = x2 + ln x. (6) Solve the D.E. below, given that y1 = e−x
2 /2
is one solution
y 00 + xy 0 + y = 0, such that y(0) = 0 and y 0 (0) = 1. (7) Prove that the equation y 00 + P (x)y 0 + Q(x)y = 0, can be transformed into
d2 y dx + Q( )2 y = 0. 2 dt dt
Hence, solve the D.E. (1 − x2 )y 00 − xy 0 + n2 y = 0. (8) By putting x = sin t solve the following D.E. q
(1 − x2 )y 00 − xy 0 + 4y = 2x (1 − x2 ).
73
Chapter 5 Series Solution 5.1
Series solution
Consider the homogeneous linear D.E. P (x)
d2 y dy + Q(x) + R(x)y = 0, 2 dx dx
(5.1)
with P (x) 6= 0. If P (x), Q(x) and R(x) are polynomial in x, in this case, the form of the D.E. suggests that we guess a polynomial solution of y. Thus, in principle we can determine a polynomial solution of y(x) by setting the sums of the coefficients of like powers of x in (5.1) to zero. Example. Find the general solution of the D.E. y 00 − 2xy 0 − 2y = 0. We set y = a0 + a1 x + a2 x2 + a3 x3 + .. =
∞ X
an xn
n=0
Hence, y0 =
∞ X
nan xn−1 , y 00 =
n=0
∞ X n=0
74
n(n − 1)an xn−2
(5.2)
so that the D.E. becomes ∞ X
n(n − 1)an xn−2 − 2x
n=0
or
∞ X
nan xn−1 − 2
n=0
∞ X
n(n − 1)an xn−2 − 2
n=0
∞ X
a n xn = 0
n=0
∞ X
nan xn − 2
n=0
∞ X
an xn = 0.
(5.3)
n=0
We can now factorize xn but the first term needs some amendments, i.e. ∞ X
n(n − 1)an xn−2 =
n=0
∞ X
(n + 2)(n + 1)an+2 xn
n=−2
where we have replace n by n + 2 in all terms without affecting the summation, but the limits changes from 0 → ∞ to n = −2 → ∞. However, one can write ∞ X
(n + 2)(n + 1)an+2 xn =
n=−2
∞ X
(n + 2)(n + 1)an+2 xn
n=0
since the summation over n = −1 and n = −2 are zero !. Therefore, equation (5.3) becomes [
∞ X
∞ X
(n + 2)(n + 1)an+2 − 2
n=0
nan − 2
n=0
∞ X
an ]xn = 0
n=0
which has the solution ∞ X
(n + 2)(n + 1)an+2 − 2
n=0
∞ X n=0
nan − 2
∞ X
an = 0
n=0
which yields
2 an , n = 0, 1, 2, ... . n+2 Setting n = 0 we obtain a2 = a0 , n = 1 gives a3 = 23 a1 and a4 = 12 a0 , a5 = 52 a3 = 22 a , a = 13 a4 = 12 13 a0 , .... 53 1 6 an+2 =
We see that one can write all the coefficients of the series (5.2) in terms of only a0 and a1 . Note that y 0 (x) = a1 + 2a2 x + 3a3 x2 + .... 75
together with the series (5.2) gives y(0) = a0 and y 0 (0) = a1 . Thus a0 and a1 must be arbitrary until specific conditions (e.g. y(0), y 0 (0)) are imposed on y. To find two solutions of our original D.E., we choose two different set of values of a0 and a1 . The simplest possible choices are (i) a0 = 1, a1 = 0, (ii) a0 = 0, a1 = 1. (i) a0 = 1, a1 = 0 In this case a3 = 0, a5 = 0, ... i.e., all the odd coefficients are zero since a3 = 2 a ,a 3 1 5
= 25 a3 , a7 = 27 a5 and so on. The even coefficients are determined from the
relations
1 1 1 a2 = a0 =, a4 = a2 , a6 = a4 = a0 , 2 3 2.3 and so on. By induction one can write for any even coefficients (a2n ) a2n =
1 1 = . 1.2.3.4...n n!
Hence, the first solution is y1 (x) = 1 + x2 + since ex = 1 + x +
x2 2!
+
x3 3!
+
x4 4!
x4 2!
+
x6 3!
+ ... =
P∞
x2n n=0 n! 2
2
= ex ,
+ .. (and here we replace x by x ).
(ii) a0 = 0, a1 = 1 In this case all even coefficients are zero, and the odd terms are defined by 2 2 2 22 2 222 a3 = a1 = , a5 = a3 = , a 7 = a5 = 3 3 5 53 7 753 and by induction one can write for any odd coefficient (a2n+1 , since 2n + 1 is always an odd number) a2n+1 =
2n . 3.5.7....(2n + 1)
2
2 Thus y2 (x) = x + 23 x3 + 3.5 x5 + .. is the second solution. Now observe that y1 (x), y2 (x) are polynomials of infinite degree even though the
coefficients P (x) = 1, Q(x) = −2x and R(x) = −2 are polynomials of finite degree. Such polynomials are called Power Series. Many of the functions f (x) that arise in applications can be expanded in power series; that is, we can find coefficients a0 , a1 , a2 , a3 , ... so that 76
P
n f (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + a3 (x − x0 )3 + ... = ∞ n=0 an (x − x0 ) . Such functions are said to be analytic at x = x0 , and the series is called Taylor
Series of f (x) about x = x0 . Exercise.
n
Show that if f (x) admits an expansion, then an =
( ddxnf )|x=x0 . n!
Examples. (1) Find two linearly independent solutions of y 00 +
3x 0 1 y + 2 y = 0. 2 1+x x +1
(5.4)
This can be written as (1 + x2 )y 00 + 3xy 0 + y = 0. Setting y=
∞ X
an xn , so that,
y0 =
n=0
∞ X
nan xn−1 , y 00 =
n=0
∞ X
n(n − 1)an xn−2 .
n=0
Hence, equation (5.4) becomes ∞ X
n(n − 1)an xn−2 + x2
n=0
∞ X
n(n − 1)an xn−2 + 3x
n=0
∞ X
nan xn−1 +
n=0
∞ X
an xn = 0,
n=0
or ∞ X
n(n − 1)an x
n−2
+
n=0
∞ X
n
n(n − 1)an x + 3
n=0
∞ X n=0
n
nan x +
∞ X
an xn = 0. (5.5)
n=0
We see that the term xn can be made a common factor but the first term needs some amendments, i.e., ∞ X n=0
n(n − 1)an x
n−2
=
∞ X n=−2
77
(n + 2)(n + 1)an+2 xn ,
where we have changed n by n + 2 every where in the summation. However, the sum for n = −2, n = −1 gives zero and hence, one can write ∞ X
(n + 2)(n + 1)an+2 =
n=−2
∞ X
(n + 2)(n + 1)an+2 .
n=0
Hence, equation (5.5) yields (
∞ X
(n + 2)(n + 1)an+2 +
n=0
∞ X
n(n − 1)an + 3
n=0
∞ X
nan +
n=0
∞ X
an )xn = 0,
n=0
which implies that the bracket is zero, i.e., an+2 = −
n+1 an , n = 0, 1, 2, ... n+2
Such a relation is called a recurrence formula for the coefficients a2 , a3 , ... in terms of a0 and a1 . To find two linearly independent solutions of equation (5.4), we choose the two simplest cases: (i) a0 = 1, a1 = 0 In this case, all odd coefficients are zero since a3 = − 23 a1 = 0, a5 = − 45 a3 = 0, and so on. The even coefficients are determined from the relations 1 1 3 1.3 5 1.3.5 a2 = − a0 = − , a4 = − a2 = , a6 = − a4 = − , 2 2 4 2.4 6 2.4.6 and so on, and by induction one can write, for any even coefficient (a2n ) a2n = (−1)n Thus, y1 (x) = 1 − 12 x2 +
1.3.5...(2n − 1) 1.3.5...(2n − 1) = (−1)n . 2.4.6...2n 2n n! 1.3 4 x 2.4
+ ... is one solution.
(ii) a0 = 0, a1 = 1 In this case, all even coefficients are zero, and all odd coefficients can be obtained from the relations: 2 2 4 2.4 6 2.4.6 a3 = − a1 = − , a5 = − a3 = , a7 = − a5 = − , 3 3 5 3.5 7 3.5.7 78
and so on, and by induction one can write for any odd coefficient the relation (a2n+1 ) 2.4.6...2n 2n n! a2n+1 = (−1)n = (−1)n . 3.5.7...(2n + 1) 3.5...(2n + 1) Thus, y2 (x) = x − 23 x3 +
2.4 5 x 3.5
+ ... is the second solution.
(2) Solve y 00 + x2 y 0 + 2xy = 0,
with y(0) = 1, y 0 (0) = 0.
(5.6)
Setting y=
∞ X
n
0
an x , so that, y =
n=0
∞ X
00
, y =
n=0 ∞ X
n(n − 1)an x
n−2
+x
2
n=0
or
nan x
n−1
∞ X
∞ X
n=0
∞ X
n(n − 1)an xn−2 .
n=0
nan x
n−1
∞ X
+ 2x
n=0
n(n − 1)an xn−2 +
∞ X
an xn = 0.
n=0
nan xn+1 + 2
n=0
∞ X
an xn+1 = 0.
(5.7)
n=0
We see that the term xn+1 can be made a common factor but the first term needs some amendments, i.e., ∞ X
n(n − 1)an xn−2 =
n=0
∞ X
(n + 3)(n + 2)an+3 xn+1
n=−3
where we have changed n by n + 3 every where in the summation. However, the sum for n = −3, n = −2 gives zero (but n = −1 gives 2a2 ) and hence, one can write
∞ X
∞ X
(n + 3)(n + 2)an+3 = 2a2 +
n=−2
(n + 3)(n + 2)an+3 .
n=0
Hence, equation (5.7) yields 2a2 + (
∞ X
(n + 3)(n + 2)an+3 +
n=0
∞ X n=0
79
nan + 2
∞ X n=0
an )xn+1 = 0,
implies that the bracket is zero, i.e., an+3 = −
1 an , n = 0, 1, 2, ... n+3
Setting the sums of the coefficients of like powers of x equal to zero gives 2a2 = 0, a3 = − 13 a0 , a4 = − 14 a1 , a5 = − 15 a2 , a6 = − 61 a3 , a7 = − 17 a4 , a8 = − 18 a5 , a9 = − 19 a6 , but since a2 = 0, a5 = 0, a8 = 0, a11 = 0, ... and a3 = − 13 a0 , a4 = − 14 a1 , a6 = − 16 a3 = 16 13 a0 , a7 = 17 41 a1 , ..., and by induction one can write for any odd coefficients a3n =
(−1)n (−1)n (−1)n a0 = n a0 = n a0 . 3.6...3n 3 .2.3.4 3 n!
The value of the coefficient a0 and a1 are determined from y(0) = a0 , y 0 (0) = a1 , hence, a0 = 1, a1 = 0 (see equation (5.2)). Hence, ∞ X 1 x6 x9 (−1)n 3n y(x) = 1 − x3 + − + ... = x . n 3 3.6 3.6.9 n=0 3 n!
(3) Solve the D.E. (x2 − 2x)y 00 + 5(x − 1)y 0 + 3y = 0, with y(1) = 7, y 0 (1) = 3 Since the initial condition are given at x = 1, we express the coefficients of the D.E. above as polynomials in (x − 1), and we find y(x) as a power series centered about x = 1. To this end, we observe that x2 −2x = x(x−2) = [x−1+1][(x−1−1)] = [(x−1)+1][(x−1)−1)] = [(x−1)2 −1]. Hence, one can write the D.E. as [(x − 1)2 − 1]y 00 + 5(x − 1)y 0 + 3y = 0.
(5.8)
Setting y=
∞ X n=0
an (x−1)n , so that, y 0 =
∞ X
nan (x−1)n−1 , y 00 =
n=0
∞ X n=0
80
n(n−1)an (x−1)n−2 .
Hence, equation (5.8) yields [(x−1)2 −1]
∞ X
n(n−1)an (x−1)n−2 +5(x−1)
n=0
∞ X
nan (x−1)n−1 +3
n=0
∞ X
an (x−1)n−1 = 0,
n=0
or ∞ X
n(n−1)an (x−1)n −
n=0
∞ X
n(n−1)an (x−1)n−2 +5
n=0
∞ X
nan (x−1)n +3
n=0
∞ X n=0
an (x−1)n = 0, (5.9)
n
We factorize (x − 1) , but the second term should be changed as ∞ X
n(n − 1)an (x − 1)n−2 =
n=0
∞ X
(n + 2)(n + 1)an+2 (x − 1)n ,
n=−2
=
∞ X
(n + 2)(n + 1)an+2 (x − 1)n ,
n=0
since the summation for n = −2 and n = −1 give zero contribution. Hence, equation (5.9) gives an+2 =
n+3 an , n+2
n = 0, 1, 2, ... .
The initial conditions determine a0 , a1 , which give a0 = y(1) = 7, a1 = y 0 (1) = 3 or a0 = 7 and a1 = 3. We see that a2 = 23 a0 , a3 = 43 a1 , a4 = 54 a2 = 6 a 5 3
=
64 a ,a 53 1 6
= 76 a4 =
a2n =
753 a, 642 0
5.3 a ,a 4.2 0 5
=
and by induction one can write for any coefficients
3.5....(2n + 1) .7, 2.4.6...2n
and a2n+1 =
4.6....(2n + 2) .3 3.5.7.(2n + 1)
Hence, y(x) = 7+3(x−1)+ 32 .7(x−1)2 + 43 .3(x−1)3 +... is the required solution. Exercise 5.1 Solve the following D.E. with the given initial conditions (i) y 00 + x2 y 0 = 0, y(0) = 2, y 0 (0) = −1, (ii) y 00 + y 0 + xy = 0, y(0) = −1, y 0 (0) = 2, (iii) y 00 + x3 y 0 + 3x2 y = 0, y(0) = 2, y 0 (0) = −1, (iv) y 00 − x3 y 0 = 0, y(0) = 0, y 0 (0) = −2. 81
5.2
Singular points, the method of Frobenius
The D.E. P (x)y 00 + Q(x)y 0 + R(x)y = 0 is said to be regular at x = x0 , if P (x0 ) = 0. There are certain special cases where we can find solutions y(x) of the form y(x) = (x − x0 )r [a0 + a1 (x − x0 ) + ....] =
∞ X
an (x − x0 )n+r ,
n=0
where n is an integer but r need not be an integer. Theorem: Now consider P (x)y 00 + Q(x)y 0 + R(x)y = 0, x ≥ x0 ,
(5.10)
and assume that x = x0 is a singular point of equation (5.10). For simplicity we set x0 = 0. It is possible, sometimes, to solve equation (5.10) by the method x2 R(x) P (x)
to be analytic at x = x0 .
Q(x) b0 = + b1 + b2 x + b3 x2 + ... P (x) x
(5.10a)
of Frobenius. This requires that Equivalently,
xQ(x) P (x)
and
and c0 c1 R(x) = 2 + + c2 + c3 x + ... P (x) x x
(5.10b)
In this case we say that x = x0 is a regular singular point of equation (5.10). Then, we can always find at least one solution y(x) of the form r
2
3
y(x) = x [a0 + a1 x + a2 x + a3 x + ...] =
∞ X
an xn+r
n=0
for some r (possibly complex) [xr = xα+iβ = xα xiβ = xα eiβ ln x ]. Now xQ(x) P (x)
= b0 + b1 x + b2 x2 + b3 x3 + ... and 82
x2 R(x) P (x)
= c0 + c1 x + c2 x2 + c3 x3 + ...
will converge for |x| < ρ. Now substitute these in equation (5.10) to obtain the equation r2 + (b0 − 1)r + c0 = 0, which is called the indicial equation with r1 ≥ r2 if they are real, where r1 and r2 are its roots. Then the D.E. has two linearly independent solutions y1 and y2 on the interval 0 < x < ρ of the following form1 : (a) if r1 − r2 is not a positive integer, then y1 = xr1
∞ X
an xn
and y2 = xr2
n=0
∞ X
b n xn .
n=0
(b) if r1 = r2 , then y1 = x
r1
∞ X
an x
n
and y2 = y1 ln x + x
n=0
∞ X
r1
bn xn .
n=0
(c) if r1 − r2 = N is a positive integer, then y1 = xr1
∞ X
an xn
and y2 = ay1 ln x + xr2
n=0
∞ X
b n xn .
n=0
The correct a may turn out to be zero. Examples. (1) Find the general solution of the D.E. x2 y 00 − 2xy 0 + 2y = 0. We have (see equations (5.10), (5.10a) and (5.10b)) b0 = −2 (the coefficient of x1 ), c0 = 2 (the coefficient of
1 ), x2
so that the indicial equation r2 + (b0 − 1)r + c0 = 0,
or r2 − 3r + 2 = 0, i.e., r = 1, 2. The general solution can then be obtained (see (a)-(c) above). (2) Find the indicial equation, its roots, and the general solution of the following D.Es. (a) y 00 − 1
1 0 1 + x2 00 2 0 00 3 0 1 + x y+ y = 0, (b) y + y +xy = 0, (c) y + y + 2 y = 0. 2x 2x2 x x x
see page 108
83
(a): We see that b0 = − 12 , c0 = 21 , so that the indicial equation r2 − 23 r + yields r1 = 1, r2 = x
P∞
n=0
a n xn , y 2 = x
1 2
1 , 2
1 2
= 0,
hence r1 − r2 is not a positive integer. Thus, y1 =
P∞
n=0 bn x
n
can be found.
(b): We have b0 = 2, c0 = 0 so that the indicial equation r2 + r = 0 yields P n r1 = 0, r2 = −1, so r1 − r2 is a positive integer. Therefore, y1 = ∞ n=0 an x , y2 = x−1
P∞
n=0 bn x
n
or logarithmic.
(c): Here we have, b0 = 3, c0 = 1 so that the indicial equation r2 + 2r + 1 = 0 yields r1 = −1, r2 = −1, i.e., r1 = r2 . This is a logarithmic case and y1 = P
P
∞ n −1 n x−1 ∞ n=0 an x , y2 = y1 a ln x + x n=0 bn x . (3) Find two linearly independent solutions of the D.E.
2xy 00 + y 0 + xy = 0, Let y =
P∞
n=0
(5.11)
an xn+r , a0 6= 0. Compute
∞ X
y0 =
0 0: In this case, m1 and m2 are negative, and the solution of equation (6.3) has the form, x(t) = c1 em1 t + c2 em2 t .
(6.5)
(ii) c2 − 4km = 0 (equal roots): In this case the solution is represented by x = c1 e−ct/2m + c2 te−ct/2m .
(6.6)
(iii) c2 − 4km < 0: In this case we have complex roots. Thus, x = e−ct/2m (c1 sin βt + c2 cos βt), where β =
√ 4km−c2 . 2m
(6.7)
The first two cases referred to as overdamped and critically
damped motion, respectively. The third case, which is referred to as underdamped motion, happens quite often in mechanical systems. This is because in all cases x involves a term that decreases rapidly with time t. (2) Free vibration: This is the simplest case of free undamped motion. In this case, equation (6.3) reduces to
d2 x k + x = 0. 2 dt m
(6.8)
Or
d2 x + ω02 x = 0, dt2 k where ω02 = m . The general solution of equation (6.9) is x(t) = A sin ω0 t + B cos ω0 t,
(6.9)
(6.10)
where A and B are constants. (3) Forced free vibration: We now remove the damping term from our system and consider the case of forced
97
free vibration, where the forcing term is periodic and has the form F = F0 cos ωt. In this case the D.E. governing the motion of the mass m is given by d2 x F0 + ω02 x = cos ωt. 2 dt m
(6.11)
In this case the general solution of the D.E. yields x(t) = c1 cos ω0 t + c2 sin ω0 t +
F0 cos ωt. m(ω02 − ω 2 )
(6.12)
Now the very interesting case is when ω = ω0 ; that is, when the frequency ω of the external force equals the natural frequency of the system. This case is called the resonance case, and the D.E. of the motion for the mass m is F0 d2 x + ω02 x = cos ω0 t. 2 dt m
(6.13)
We see that the coefficient of the cosine term (the amplitude) in equation (6.12) becomes infinitely large as ω → ω0 ! Such a phenomenon was responsible for the collapse of the Tacoma Bridge (USA) in 1940.
6.2
Schr¨ odinger Equation
This equation describes the motion of micro-particles (e.g. electron, proton, neutron, etc.). It is a second order D.E., it has the form d2 ψ 2m + 2 (E − V )ψ = 0, dx2 h ¯
or ψ 00 +
2m (E − V)ψ = 0 h ¯2
(6.14)
where E = constant, is the total energy of the particle of mass m, V is the potential energy of the particle, h = 2π¯h is known as Planck’s constant. ψ is called the wave function of the particle, since micro-particles have wave nature. Thus, the solution ψ depends only on the form of the potential V (x).
98
Examples: (1) V (x) = constant, in this case Schr¨odinger equation reads, ψ 00 + k 2 ψ = 0, where k 2 =
2m (E h2 ¯
− V )=const., thus one has, if E > V , ψ = c1 sin(kx) + c2 cos(kx),
and, if E < V , ψ = c1 ekx + c2 e−kx . (2) V (x) = 12 kx2 , then one gets the solution of a simple harmonic motion (quantum mechanically). Such a system represents the vibrational motion of atoms in a solid crystal. The resulting D.E. can be written as, ψ 00 + aψ − bx2 ψ = 0, a , b, const.. In this case the solution is obtained by the method of series solution discussed previously (see exercise (7) page 93).
6.3
Electric circuit
Consider an electric circuit consisting of a capacitor (C), an inductor (L), a resistor (R) and a power generator (E). The total voltage is the sum of the voltages of each component. Therefore, E(t) = L
dI Q + IR + , dt C
where L dI is the voltage drop across the inductor (L is called the inductance), dt IR is the voltage drop across the resistor and Q , is the voltage drop across the C capacitor (C is called capacitance, Q is the charge). 99
The above equation can be written as (since I = L
dQ ) dt
d2 Q dQ Q +R + = E(t), 2 dt dt C
which is a non-homogeneous second order D.E. with constant coefficients (L, R, C). The solution of such an equation is already given in sec.4.2. As discussed previously, this system has a resemblance with the mechanical system investigated in sec.6.1. For example, the tuning knob of a radio is used to vary the capacitance in the tuning circuit. In this manner, the resonant frequency is changed until it coincides with the frequency with one of the incoming radio signals. In this case the amplitude of the current produced by this signal will be the maximum one in relation to other signals. In this way, the tuning circuit picks out the desired station. Examples: (1) Consider a C-L circuit (i.e., R = 0) then our D.E. becomes ¨+ LQ
Q = E0 sin ωt. C
This is a D.E. and can be written as ¨ + Q = E0 sin ωt. Q LC L Hence, (D =
d ) dt
(D2 +
E0 1 )Q = sin ωt. LC L 100
The complementary solution is Qc = A sin √
1 1 t + B cos √ t, LC LC
and the particular solution is Qp =
E0 sin ωt L 1 2 D + LC
=
E0 L
sin ωt 1 . + LC
−ω 2
The general solution is E0 sin ωt 1 1 Q = A sin √ t + B cos √ t+ L 2 1 . −ω + LC LC LC
(2) Solve the following simultaneous D.Es. dx 1 dy 3 + 18x + + y + 5e2t = 0, dt 2 dt 2 dx dy + 22x + + 2y + 7e−2t = 0. dt dt Here x, y are dependent variables and t is the independent variable. We have to find x, y in terms of t and possible integration constants. First write the equations 5
in the forms
1 (D + 18)x + (D + 3)y = −5e2t , 2 (5D + 22)x + (D + 2)y = −7e−2t ,
(6.15) (6.16)
where D = dtd . Now operate with 2(D + 2) on equation (6.15) and with (D + 3) on equation (6.16) and subtract to eliminate y and its derivative. Then [2(D + 2)(D + 18) − (D + 3)(5D + 22)] = −10(D + 2)e2t + 7(D + 3)e−2t , or
10 2t 7 −2t e − e . 3 3 The complementary solution of the above equation is (D − 2)(D + 1)x =
xc = Ae2t + Be−t . 101
(6.17)
The particular solution is obtained from xp =
− 73 e−2t . (D − 2)(D + 1) 10 2t e 3
40 2t 7 te − e−2t . (6.18) 9 12 To find y, we now eliminate Dy from equations (6.15), (6.16) to obtain y directly in terms of x, Dx and know function of t. To this end, multiply equation (6.15) xp = Ae2t + Be−t +
by 2 and subtract equation (6.16) to give (−3D + 14)x + y = −10e2t + 7e−2t , yp = (3D − 14)xp − 10e2t + 7e−2t . This gives yp = −8Ae2t − 17Be−t + (
10 320 2t 56 −2t − t)e + e . 3 9 3
This equation finds an application in electric circuit (a transformer circuit). Exercise 6.1 (1)-(a) In a C-L circuit, find the resonant frequency of the system in terms of C, L; if the driving voltage is E = E0 sin ωt. (1)-(b) If the circuit does not contain an inductor and E = 0, find the time at which the charge (Q) reduces to
1 e
of its initial value.
(2) Find the equation of motion of a simple pendulum subjected to viscous drag ˙ where θ is angle between the pendulum string and the vertical, k =const., = −k`θ, ` is the length of the string, [consider a small vibration of the pendulum] (3) Let a particle of mass m move on the x-axis subject to a force F = −kx+g(v), where k is a positive constant and g(v) = −bv 2 , b is a constant and v is the velocity. Find v in terms of x and give a physical interpretation. (4) Let a particle of mass m move on the x-axis subject to a force F ; let x = x0 , v = v0 when t = 0. Find x and v in terms of t for each of the following 102
cases (m, g, b, and k are constants): (a) F = −mg − bv , (b) F = −bv + k sin t , (5) Solve the following simultaneous D.Es. 3
dx dy + − x + y = 4 cos t, dt dt
dx dy + − 2x = e−t , dt dt given that at t = 0, x = 0, y = 0. (6) Solve the following simultaneous D.Es. dx dy + 2 − x − 5y = sin 2t + cos 2t, dt dt dx dy + 3 − 2x − 6y = 9et , dt dt given that at t = 0, x = 0, y = 0.
103
References Zalman Rubinstein, A course in ordinary and partial differential equations, Academic Press(U.K.),1969 M. Braun, Differential equations and their applications, Springer-Verlag (U.S.A.), 1975 Frank Ayres, Jr., Differential equations, Schaum’s outline series, McGraw-Hill Book Company (1952) Wilfred Kaplan, Ordinary differential equations, Addison-Wesely Publishing Company (USA), 1958 Frank Chorlton, Ordinary differential and differential equations, D.Van Nostrand Company LTD.(UK), 1965 R. A. Agnew, Differential Equations, McGraw-Hill Book Company (1960) David. A. Sanchez. Differential equations, Addison-Wesely Publishing Company (USA), 1988
104
6.4 R
Table of Integration xn dx =
xn+1 n+1
+ C, n 6= −1
R dx = ln |x| + C x R x x a dx = lna a + C , a > 0 R ax 1 ax R R R R
e
dx = a e
cos ax dx =
+C
1 a
sin ax + C
sin ax dx = − a1 cos ax + C csc2 x dx = − cot x + C
sec2 x dx = tan x + C R √ dx = sin−1 xa + C a2 −x2 R R R R R R R R
dx = a1 tan−1 xa + C a2 +x2 dx 1 = 2a ln x−a +C x2 −a2 x+a −1 dx x √ = cosh a + C x2 −a2 √ dx = ln (x + x2 ± a2 ) 2 2 x ±a
+C
cosh x dx = sinh x + C sinh x dx = cosh x + C dx 1+cos x 2
= tan x2 + C
tan x dx = tan x − x + C
R dx = ln | tan x2 | + C sin x
Taylor expansion of some common functions ex = 1 + x + 2!1 x2 + 3!1 x3 + .... 1/(1 − x) = 1 + x + x2 + x3 + x4 + ..., |x| < 1 ln(1 − x) = −(x + sin x = x − cos x = 1 −
1 3 x 3! 1 2 x 2!
+ +
3 4 x2 + x3 + x4 2 1 5 x − .... 5! 1 4 x + .... 4!
+ ...), |x| < 1
105
Convergence of a series In obtaining the Frobenius series, it is important to know the range of the values of x for which the series converges. To this end, we summarize the ratio test for convergence. If each term of the series u0 + u1 + u2 + ... + un + ... is positive (i.e., ur > 0 (r = 1, 2, 3, ...)), and if further lim n→∞
un+1 = ` < 1, un
then the series converges. When, however, we have a power series of the form a0 + a1 x + a2 x2 + ... + an xn + ..., the convergence is studied in the following way. Write un = an xn (n = 0, 1, 2, ...). Then |
un+1 an+1 |=| x | . | |. un an
Now impose lim | n→∞
an |→ R as n → ∞. an+1
Then lim | n→∞
un+1 |=| x | /R. un
R is called the radius of convergence. The series converges for |x| < R and diverges for |x| > R. Let us suppose | x |< R. Then, lim |
n→∞
un+1 |< 1, un
so that by the ratio test the positive series | u0 | + | u1 | + | u2 | +...+ | un | +... 106
is convergent. This implies that the series u0 + u1 + u2 + ... + un + ... is absolutely convergent and hence also convergent for all values of x satisfying |x| < R. Example: 1
Consider y = a0 x 2 (1 − 3!1 x2 + 5!1 x4 + ...). We have to discus the convergence of (1 − 3!1 x2 + 5!1 x4 + ...). Write un = lim | n→∞
(−1)n x2n (2n+1)!
= a2n x2n , where a2n =
(−1)n . (2n+1)!
Then
a2n+1 1 |= n→∞ lim = 0. a2n (2n + 2)(2n + 3)
Regular Singular Points Consider the equation P (x)y 00 + Q(x)y 0 + R(x)y = 0, P
P
P
n n n R(x) = ∞ Q(x) = ∞ where P (x) = ∞ n=0 cn x . This n=0 bn x , n=0 an x , equation is said to have a regular singular point at x = 0 if, after cancellation of
the highest power of x which is a common factor of the coefficients, the equation takes the form (i) (a1 x + a2 x2 + ...)y 00 + (b0 + ....)y 0 + (c0 + ...)y = 0,
(a1 6= 0),
(ii) (a2 x2 + a3 x3 + ...)y 00 + (b1 x + ....)y 0 + (c0 + ...)y = 0, (a2 6= 0). When P (x0 ) 6= 0, we say that x0 is an ordinary point. Example: Consider the D.E. x2 y + xy 0 + (x3 − 1)y = 0. Here we have P (x) = x2 , i.e. a0 = 0, a1 = 0, a2 = 1, a3 = a4 = ... = 0(a2 = 6 0), Q(x) = x, i.e. b0 = 0, b1 = 1, b2 = b3 = ... = 0, R(x) = x3 − 1, i.e. c0 = −1, c1 = 0, c2 = 0, c3 = 1, c4 = c5 = ... = 0 so that (ii) applies (our D.E. above can be written as equation (ii)) and we have a regular singular point at x = 0 107
Theorem Consider a second order D.E. of the form L(y) = 0, at a regular singular point at x = 0: where, L = P (x)D2 + Q(x)D + R(x), L(n2 ) = f (n)xn+h + g(n)xn+h+k , f (n) of degree 2, k > 0, f (r) = 0 for r = r1 , r2 , φ(x, r) = xr (1 +
∞ X
(−1)s xks
s=1
φ1 (x, r) =
g(r)...g[r + (s − 1)k] ), f (r + k)...f (r + sk) ∂φ , ∂r
ψ(x, r) = (r − r1 )φ(x, r), ψ1 (x, r) =
∂ψ . ∂r
(1) If (r2 − r1 )/k is not an integer or r2 = r1 + N k, g(r1 + νk) = 0, ν, N integers, 0 ≤ ν ≤ N − 1 then, y1 = φ(x, r1 ), y2 = φ(x, r2 ) (2) r1 = r2 then y1 = φ(x, r1 ), y2 = φ1 (x, r1 ) (3) r2 = r1 + N k, N is a positive integer and g(r1 + νk) 6= 0 for ν = 0, ..., N − 1 then y1 = ψ1 (x, r1 ), y2 = φ(x, r2 ). The general solution is written as y = c1 y1 + c2 y2 .
108
6.5
Answers to exercises
Exercise 1.1 (1) (i) order 1, degree 1 (ii) order 2, degree 1, (iii) order 2, degree 2 (iv) order 2, degree 1. (2) (i) y = y 0 x, (ii) y 00 + y = 0, (iii) y 00 = 0, (iv) x2 y 00 − 2xy 0 + 2y = 0, (v) y 00 (1 − x) + xy 0 − y = 0 dy dy 2 (3) (x dx − y)2 = 1 + ( dx ).
(4) (i) y 00 + y − 6 = 0, (ii) y = y 0 x − 2x2 , (iii) y 00 + y = 0 Exercise 1.2 (i) particular, (ii) general, (iii) general (iv) particular, (v) general (vi) general Exercise 1.3 (1) (i) y = C(1 + x2 )−1/2 , (ii) y 2 = Cx−2 + 12 x2 , (iii) ln x = −y 3 x−3 + C, (iv) xy = Cey/x 2+y−4x (2) (i) ln xy = xy + C, (ii) ln 2−y+4x = 4x + C
(3) (i) y ln x + x = y, (ii) y = 4x−2 , (iii) y 2 + 12 x2 = 23 x−2 (4) y 2 = 2x2 ln x + Cx2 . (5) y 3 = x−2 . (6) f (x) = −2 cos x + C, x4 + 2yx2 = C. (7) f (x) = 21 x + Cx−1 . Exercise 1.4 (1) (i) (y − x)3 = C(y + x − 2), (ii) x2 − 2x + 8y 2 − 14y = C, (iii) (y − x + 1)5 (y + x − 1) = C, (vi) 6y − 3x = ln(3y + 3x + 2) + C, (v) x − y = C − 2 ln (x − 2y + 1), (vi) 2(x − y) + ln (2y + 2x − 1) = C. Exercise 2.1 4
y x 1 5 2 (1) (i) 13 x3 − yx = C, (ii) ln x − 4x 4 = C, (iii) ln x − 2y = C, (iv) yx − 5 x = C.
(2) a = 1, x + e−x sin y = C. 109
(3) a = −2, (2x−1) − x1 = C. 2y 2 √ (4) y 1 − x2 = C + sin−1 x. Exercise 3.1 2
(1) (i) y −4 = − 12 + e3x , (ii) y = − cos x + sinx x + Cx , (iii) y = −x sec x + C sec x, (iv) y(C cos x − sin x) = cos2 x. Exercise 3.2 (i) y = Cx − 2C 2 , (ii) y =
y = x, (iii) y = C 2 − Cx−1 , y = − 41 x−2 ,
(iv) y 3 = C 2 +3Cx, (v) y
Cx−C 2 , (vii) y 2 = 2Cx+C 3 , y 4 = − 21 x3 ,
(viii) y = 21 Cx2 −
2 + 12 x2 , C = 14 x2 , y =
1 . 2C
Exercise 3.3 (1) (a) y = 2 − cos x, (b) y = ex , (c) y = ex − ex + 1. (2) (i) ex + e−y = C, (ii) y = Ce3x , (iii) y − 2 = Cex (y − 1), (iv) 4y 3 = 3x4 + C, y = 12 x2 + x + C. (v) e− cos y cos x = C, (vi) ln y+1
(3) (i) x2 − 2xy − y 2 = C, (ii) e−y/x + ln x = C, (iii) x3 + xy 3 = C, (iv) tan[ (y−x) ] = Cx. 2x Exercise 4.1 (1) all are independent W 6= 0. Exercise 4.2 (1) (i) y = c1 ex + c2 e2x + c3 e−x , (ii) y = c1 e2x + c2 xe2x , (iii) y = c1 ex + c2 e−2x , (iv) y = c1 e2x + c2 xe2x + c3 e−2x , (v) y = e−x (c1 cos x + c2 sin x). (2) (4 − 3x)ex , 2 − 3x2 , 0, 2 − 3x2 . Exercise 4.3 (ii) y = c1 e−x + c2 e−4x +
− 12 x, (iii) y = c1 ex + c2 e−x + ex (x2 − x + 12 ),
(iv) y = c1 + c2 e2x + c3 e
− 18 x2 −
11 8 −2x
1 , 16
(v) y = c1 ex + c2 e2x + c3 xe2x + 12 x2 e2x .
Exercise 4.4 (2) (i) y = c1 sin 6x + c2 cos 6x + (ii) y = c1 e2x + c2 xe2x +
2 27
sin 3x,
3 (−77 sin 9x 7225
+ 36 cos 9x), (iii) y = e−2x (c1 sin 3x +
110
c2 cos 3x) +
2 x 13
+
Exercise 4.5 (1) (i) y = c1 sin
√
7 3x e 34
−
8 , 169
2x + c2 cos
(iv) y = c1 e−2x + c2 e−3x +
√
1 x e 12
+ 12 e−x .
2x + 13 e3x + 1, (ii) y = c1 e−x + c2 xe−x − 15 (sin x −
2 cos x) + 21 x2 − x + 21 (iii) y = c1 ex + c2 e−x − 18 ex (cos 2x + 2x sin 2x + 2x cos 2x), (iv) y = c1 e2x + c2 e3x + 14 e4x (2x2 − 6x + 25), (v) c1 e3x + c2 xe3x − e3x ln x. (2) (i) y = c1 sin x + c2 cos x + 21 x sin x, (ii) y = c1 + c2 e2x + c3 e−x + 91 (x3 − 4x2 − 26 x), 3
(iii) y = c1 sin 2x+c2 cos 2x− 14 (x cos 2x), (iv) y = c1 cos x+c2 sin x−cos ln(sec x+ tan x). Exercise 4.6 (1) y = c1 x3 + c2 x3 ln x. (2) (i) y = c1 x3 + c2 x3 ln x, (ii) y = c1 x2 + c2 x−1 , (iii) y = c1 x + c2 x−3 + x2 , (iv) y = c1 sin ln x + c2 cos ln x − 21 x(ln x − 1), (v) y = c1 x2 + c2 x−1 + 12 x2 ln x., (vi) y = x−2 [c1 cos(3 ln x) + c2 sin(3 ln x)] +
1 13
ln x −
4 169
+
1 2 x. 25
Exercise 4.7 (1) y 3 = 3c1 x + c2 .
√ (2) (i) y = c1 cos x + c2 sin x − sin x ln | csc x + cot x| , (ii) y = c1 sin 5x + √ √ √ 1 c2 cos 5x − 45 , (iii) y = e− 2 x (c1 sin 215 x + c2 cos 215 x) + 16 ex , (iv) y = c1 ex + c2 e−x − (x + 1), (v) y = c1 e−x + c2 e−2x + 1 −2x e 6
1 (sin x 10
− 3 cos x), (vi) y = 43 ex +
− (x + 12 ), (vii) y = sin x + x sin x.
(3) y = c1 xn + c2 xn+1 . (5) (i) y = (c1 ex + c2 e−x − cos x) sin x, (ii) y = (c1 xn+1 + c2 x−n )ex , (c) y = (c1 + c2 sin x) exp(− sin x) − sin2 x + 4 sin x − 5. (6) W = y1 y20 − y2 y10 = 1 . (7) (i) y = c1 x + c2 x−5 , (ii) y = (3−c)x
(8) y = c1 ecx + c2 ce(3−2c) [x −
√ 1 (1+ 3) √ x 2 3
−
√ 1 (1− 3) √ x . 2 3
c ]. (3−2c)
(9) (i) y = c1 x−1 + c2 x−1 ln x, (ii) y = c1 x + c2 x ln x. 111
Exercise 4.8 (1) y2 = (x ln x + 1)(x − 1)−2 − 1, (d) y2 = − 71 x−5 . (3) (a) y2 = x ln x, (b) y2 = 51 x3 , (c) y2 = 12 x ln 1+x 1−x 1
(4) k = − 12 , y = x− 2 [c1 + c2 sin−1 (2x − 1)], (5) y = c1 x + c2 x2 + 12 ln x + 34 x2 ln x. 1 2
1 2
(6) y = c1 e− 2 x + c1 e− 2 x
R − 1 x2 e 2 dx.
(7) y = c1 cos(n sin−1 x) + c2 sin(n sin−1 x). √ (8) y = (c1 − 14 sin−1 x)(1 − 2x2 ) + 2c2 x 1 − x2 . Exercise 5.1 1 6 7 4 1 (i) y = 2−x− 12 x4 + 15 x5 + 16 x +..., (ii) y = −1+2x−x2 + 13 x3 + 24 x − 120 x5 +...,
(iii) y = 2 − x − 12 x4 + 15 x5 +
7 x8 144
+ ... , (iv) y = −2x −
1 5 x 10
−
1 x9 144
+ ....
Exercise 5.2 (1) all are regular singular points √ 1 x x2 x3 − 3.5 − 5.7 − ... (2) (i) y = a0 1 − x2 + a1 x, (ii) y = c1 (1 − x) + c2 x 2 (1 − 1.3 −3 −5 −3 5 1 5 √ ) (− )( ) ( )( )(− ) (− 1 3 2 2 2 2 x )+ c2 x[1− 3.1!2 x+ 3.5.2! x2 − 2 3.5.7.3! x3 +.. (iii) y = c1 (1+3x+x2 + 15 Exercise 5.3 (2) y = a0 (1 − x2 − 31 x4 − 15 x6 − ...) + a1 x or y = a0 [1 − 12 x ln( 1+x )] + a1 x 1−x (3) y = x2 [c1 (4) y = x
− 12
R exp(− 41 x2 ) x3
dx + c2 ]
cos x[c1 + c2 (x tan x + ln cos x)] 1
(5) k = − 12 , y = x− 2 [c1 + c2 sin−1 (2x − 1)], (6) y = c1 e2x + c2 (2x + 5) − ex 3 22 .1.3 4 x + 2 .1.3.5. x6 + ...] 5! 7! 2 3 .... n = 1, u = exp(− 12 x2 )[c3 + c4 (1 − 2!2 x2 − 24!.1 x4 − 2 6!.1.3 x6 + ...] 3 3 7 1 4 1 6 (11) y = c1 [1 + 3x2 + x4 + 17 x6 − 77 x + ...] + c2 x 2 [1 + 34 x2 − 16 x + 64 x t t 2t 2 2t 2t 2 2t 3 .e 3 .e (12) y = c1 e 3 [1 − 3e + 8.14.2.4 − ....] + c2 e− 3 [1 − 3e + 2.4.410 − ....] 8.2 2.4 2 (14) (a) 4t ddt2y + 2 dy + y = 0, dt
(7) n = 0, u = exp(− 12 x2 )[c1 + c2 x(1 + 23 x2 +
112
9 − 1792 ]
1 + (4!x1 2 ) − (6!x1 3 ) (2!x) ex (c1 + c2 x3 + ex ).
y = c1 [1 − (16) y =
1
+ ...] + c2 x− 2 [1 −
1 (3!x)
+
1 (5!x2 )
−
1 (7!x3 )
+ ...]
Exercise 6.1 1 (1) (a) ω = √LC , (b) t = RC. (2) θ¨ + g θ˙ + k θ = 0. `
2
(3) (v −
m 1 mkb−2 2 −bt/m
(4) (a) v = e
+ kb−1 x)e2bx/m = C.
(v0 +gmb−1 )−gmb−1 , x = x0 −mgb−1 t+(1−e−bt/m )b−2 (m2 g+
mbv0 ), (b) v = (b2 + m2 )−1 [[(b2 + m2 )v0 + km]e−bt/m + (kb sin t − m cos t)], x = (b3 + m2 b)−1 [[−mv0 (b2 + m2 ) − km2 ]e−bt/m − kb(b cos t + m sin t)] + x0 + (mv0 + k)/b. (5) x = 21 sin t − t cos t, y = (2t − 12 ) sin t + (1 − t) cos t − e−t (6) x = 94 e2t +
27 −2t e 4
+ 32 sin 2t − 9e−t , y = 94 e2t − 94 e−2t − 12 sin 2t.
113
6.6
Glossary
114
Glossary above accordingly across
א و
ﺃﻋﻼﻩ ﻭﻓﻘﹰﺎ ﻝـ ﻋﺒﺭ
admit
ﻴﻘﺒل
affect
ﻴﺅﺜﺭ
already
ﻤﻥ ﻗﺒل
alternatively
ﺒﻁﺭﻴﻘﺔ ﺃﺨﺭﻯ
amendment
ﺘﻌﺩﻴل
angle
ﺯﺍﻭﻴﺔ ﺃﻱ
any appear
ﻴﻅﻬﺭ
apply
ﻴﻁﺒﻕ
arbitrary
ﻏﻴﺭ ﻤﺤﺩﺩ
arising
ﻨﺎﺸﺊ
arrange
ﻴﺭﺘﺏ
assign
ﻴﻀﻊ
associated
ﻤﺭﺘﺒﻁ
assume
ﻴﻔﺭﺽ/ﻴﺄﺨﺫ
assure
ﻴﻀﻤﻥ
because
ﺒﺴﺒﺏ
become
ﻴﺼﻴﺭ/ ﻴﺼﺒﺢ
before
ﻗﺒل
below
ﺃﺴﻔل/ ﺘﺤﺕ
ﻤﺘﺄﺭﺠﺢ
bounced bracket
ﻗﻭﺱ
bridge
ﻜﺒﺭﻯ ﺒﻭﺍﺴﻁﺔ
by called
ﻴﺴﻤﻰ/ ﻴﻌﺭﻑ
case
ﺤﺎﻝﺔ
certain
ﻤﻌﻴﻨﺔ
change
ﺘﺤﻭﻴل
characteristic
ﻤﻤﻴﺯ
choice
ﺍﺨﺘﻴﺎﺭ
choose
ﻴﺨﺘﺎﺭ
circuit
ﺩﺍﺌﺭﺓ
clear
ﻭﺍﻀﺢ
coefficient
ﻤﻌﺎﻤل
coincide
ﻤﺘﻁﺎﺒﻕ
collapse
ﻴﻨﻬﺎﺭ
compare
ﻴﻘﺎﺭﻥ
comparison
ﻤﻘﺎﺭﻨﺔ
complementary ﻤﻜﻤل complete
ﻜﺎﻤل
complex
ﻤﺭﻜﺏ
compute
ﻴﺤﺴﺏ
conclude
ﻴﺨﺘﻡ
condition
ﺸﺭﻁ
drag
confirm
ﻴﺅﻜﺩ
driving
consider
ﻴﺄﺨﺫ
due to
ﺴﺤﺏ ﺠﺎﺫﺏ/ﻓﺎﻋل ﺒﺴﺒﺏ/ ﻨﺘﻴﺠﺔ ﻝـ
consist
ﻴﺘﻜﻭﻥ
eliminate
ﻴﺘﺨﻠﺹ
constant
ﺜﺎﺒﺕ
employ
ﻴﺴﺘﺨﺩﻡ
contain
ﻴﺤﻭﻯ
equal
ﻴﺴﺎﻭﻯ
continue
ﻴﺴﺘﻤﺭ
equation
ﻤﻌﺎﺩﻝﺔ ﻤﻜﺎﻓﺊ
continuous
ﻤﺴﺘﻤﺭ
equivalent
correct
ﺼﺤﻴﺢ
evaluate
correspond
ﻴﻘﺎﺒل
ﻴﺤﺴﺏ
even number ﻋﺩﺩ ﺯﻭﺠﻲ
curve
ﻤﻨﺤﻨﻰ
evident
decrease
ﻴﻨﻘﺹ
exact
define
ﻑﻴﻌﺭ
example
definition
ﺘﻌﺭﻴﻑ
except
degree
ﺩﺭﺠﺔ
exercise
denote
ﻴﺭﻤﺯ
exist
ﻴﻭﺠﺩ
dependence
ﺍﻋﺘﻤﺎﺩ
existence
ﻭﺠﻭﺩ
derivative
ﻤﺸﺘﻘﺔ
experience
ﻴﺨﻀﻊ
determine
ﻴﺤﺩﺩ
express
ﻭﺍﻀﺢ ﺘﺎﻡ ﻤﺜﺎل ﻋﺩﺍ ﺘﻤﺭﻴﻥ
ﻴﻌﺒﺭ
different
ﻤﺨﺘﻠﻑ
expression ﺘﻌﺒﻴﺭ/ ﻤﻘﺩﺍﺭ
differentiate
ﻴﻔﺎﻀل
external
discuss
ﻴﻨﺎﻗﺵ
factor
ﻋﺎﻤل
divide
ﻴﻘﺴﻡ
family
ﻤﺠﻭﻋﺔ
downward
ﺇﻝﻰ ﺃﺴﻔل
ﺨﺎﺭﺠﻲ
limit
ﻨﻬﺎﻴﺔ
linear
ﺨﻁﻰ
replace
ﻴﺒﺩل
look for
ﻴﺒﺤﺙ
represent
ﻴﻤﺜل
phenomenon
ﻅﺎﻫﺭﺓ ﻨﻘﻁﺔ
point
polynomial ﻜﺜﻴﺭﺓ ﺍﻝﺤﺩﻭﺩ ﻤﻤﻜﻥ
possible power
ﻗﺩﺭﺓ/ ﻗﻭﺓ/ ﺃُﺱ ﺴﺎﺒﻘﹰﺎ
previously proceed
ﻴﺴﺘﻤﺭ
product
ﻨﺎﺘﺞ
properly ﺒﻁﺭﻴﻘﺔ ﺼﺤﻴﺤﺔ
required
ﻤﻁﻠﻭﺏ
resemblance
ﻤﺸﺎﺒﻬﺔ
resistance
ﻤﻘﺎﻭﻤﺔ
respectively ﻋﻠﻰ ﺍﻝﺘﺘﺎﻝﻰ responsible
ﺠﺫﻭﺭ
roots said to be satisfy
ﻴﺤﻘﻕ ﻴﺒﺤﺙ
ﺨﻭﺍﺹ
seek
proportional
ﻴﺘﻨﺎﺴﺏ
separable
ﻴﺜﺒﺕ/ ﻴﺒﺭﻫﻥ
provided that ﺸﺭﻴﻁﺔ ﺃﻥ
ﻴﻘﺎل ﺒﺄﻨﻪ ﻨﻔﺱ
same
properties prove
ﻤﺴﺌﻭل
series set
ﻴﻤﻜﻥ ﻋﺯﻝﻪ
ﻤﺘﻭﺍﻝﻴﺔ/ ﻤﺘﺴﻠﺴﻠﺔ ﻴﻀﻊ
put
ﻴﻀﻊ
set up
ﻥ ﻴﻜﻭ/ﻴﻌﺩ
rapidly
ﺴﺭﻴﻌﹰﺎ
show
ﺢ ﻴﻭﻀ/ ﻥﻴﺒﻴ
readily
ﺒﺴﻬﻭﻝﺔ
reduce to refer to
ﻴﺅﻭل ﺇﻝﻰ ﻴﺭﺠﻊ ﺇﻝﻰ
simplest
ﺃﺒﺴﻁ
since
ﺒﻡ ﺃﻥ
single
ﻤﻨﻔﺭﺩ
region
ﻤﻨﻁﻘﺔ
situation
relation
ﻋﻼﻗﺔ
so forth
ﻭﻫﻜﺫﺍ
remove
ﻴﺒﻌﺩ
solution
ﺤل
solve
ﻴﺤل
turn out
ﻴﺘﻀﺢ
some
ﺒﻌﺽ
special specific
ﺨﺎﺹ ﻤﺤﺩﺩ/ ﻤﻌﻴﻥ
type
ﻭﻀﻊ/ ﺤﺎﻝﺔ
ﻨﻭﻉ
undetermined ﻏﻴﺭ ﻤﺤﺩﺩ unit
ﻭﺤﺩﺓ
standard stop
ﻗﻴﺎﺴﻲ
until
ﻴﻘﻑ
use
ﺤﺘﻰ ﻴﺴﺘﻌﻤل
subjected
ﺨﺎﻀﻊ
usual
ﻋﺎﺩﻯ
substitute
ﺽﻴﻌﻭ
value
ﻗﻴﻤﺔ
such sufficient
ﻤﺜل ﻜﺎﻓﻲ
variable
ﻤﺘﻐﻴﺭ
vary
ﻴﻐﻴﺭ
suggest
ﻴﻘﺘﺭﺡ
verify
suitable
ﻤﻨﺎﺴﺏ
vertical
ﺭﺃﺴﻲ
sum
ﻤﺠﻤﻭﻉ
vertically
ﺭﺃﺴﻴﹰﺎ
suppose
ﻴﻔﺘﺭﺽ
vibration
ﺍﻫﺘﺯﺍﺯ
system
ﻤﻨﻅﻭﻤﺔ
viscous
term
ﺤﺩ
wave
ﻴﺘﺤﻘﻕ ﻤﻥ
ﻝﺯﺝ ﻤﻭﺠﺔ
ﺇﺫﹰﺍ/ ﺇﺫﻥ
which
ﺍﻝﺘﻰ
though
ﺭﻏﻡ
while
ﺒﻴﻨﻤﺎ
thus
ﻫﻜﺫﺍ
with respect to ﺒﺎﻝﻨﺴﺒﺔ ﻝـ
therefore
together
ﻤﻊ
without
ﺒﺩﻭﻥ
total
ﻜﻠﻰ
write
ﻴﻜﺘﺏ
yield
ﻴﻌﻁﻰ/ ﻴﻨﺘﺞ
transformation ﺘﺤﻭﻴل true
ﺼﺤﻴﺢ
