Solution 4.2(a). Use partial fractions, i.e. set s s 3 s 5. A s 3. B s 5 s. A s. 5. B s. 3.
Using the cover−up method for example we see that. A. 3. 2. , B. 5. 2 . Hence.
Chapter 4 (Laplace transforms): Solutions (The table of Laplace transforms is used throughout.)
Solution 4.1(a) 1 y j À Hsin H4 tL cos H2 tLL = À i z j sin H4 tLz { k2 1 = À H sin H4 tLL 2 1 4 = 2 2 s + 16 2 = . 2 s + 16
Solution 4.1(b) 2 y ii1 z t -t y z j z j À H cosh 2 HtL L = À j H ã ã L z j z j z j 2 { { kk
1 ã-2 t y i ã2 t j z =Àj j + + z z 2 4 { k 4
1 1 1 1 1 1 = × + × + × 4 s - 2 2 s 4 s + 2 s2 - 2 = . s Hs2 - 4L
Solution 4.1(c) 1 y j À Hcos Ha tL sinh Ha tLL = À i z j Hãa t - ã-a t L cos Ha tLz { k2 1 = À Hãa t cos Ha tL - ã-a t cos Ha tL L 2 1 i s-a s+a j = j - 2 2 2 j Hs aL + a Hs + aL2 + a2 k a s2 - 2 a3 = . s4 + 4 a4
Solution 4.1(d) 2 À Ht2 ã-3 t L = À Hã-3 t t2 L = . Hs + 3L3
y z z z {
Solution 4.2(a) Use partial fractions, i.e. set s A B = + Hs + 3L Hs + 5L Hs + 3L Hs + 5L
s = A Hs + 5L + B Hs + 3L
Using the cover−up method for example we see that 3 5 A = - , B = . 2 2
Hence s 3 1 5 1 = - × + × Hs + 3L Hs + 5L 2 Hs + 3L 2 Hs + 5L
5 3 y+Ài y. j j =Ài z j ã-5 t z j - ã-3 t z z { k 2 k 2 {
And so the inverse Laplace transform s i À-1 j j k Hs + 3L Hs + 5L
3 5 y z z = - ã-3 t + ã-5 t . 2 2 {
Solution 4.2(b) Use partial fractions, set 1 A Bs+C = + 2 2 s Hs + k L s s2 + k2
1 = A Hs2 + k2 L + HB s + CL s
1 = A s2 + A k2 + B s2 + C s
1 = HA + BL s2 + C s + A k2
Equating coefficients of s2 :
A+B=0
s1 :
C=0
s0 :
1 A = k2
1 Using the first equation we see that B = - k2 , and hence
1 1 1 1 s = × - × 2 2 2 2 2 s Hs + k L k s k Hs + k2 L s 1 i1 y z = j j - z 2 2 Hs + k2 L { k ks
1 = À H1 - cos Hk tLL. k2
And so the inverse Laplace transform 1 1 i y z H1 - cos Hk tLL. À-1 j j z = 2 2 k2 k s Hs + k L {
Solution 4.2(c) 1 , note that from the table of To find the inverse Laplace transform of Hs + 3L2 1 Laplace transforms, the Laplace transform of t is s 2 , and so if we apply the 1 -3 t . Hence shift theorem, the Laplace transform of ã t must be Hs + 3L2
1 i y -3 t j z z . À-1 j j z=tã 2 k Hs + 3L {
Solution 4.3 To solve the initial value problem y ’’ + y = t, y H0L = 0, y ’ H0L = 2;
we take the Laplace transform of both sides of the differential equation À Hy ’’ HtL + y HtLL = À HtL
À Hy ’’ HtLL + À Hy HtLL = À HtL
1 s2 y HsL - s y H0L - y ’ H0L + y HsL = s2 1 s2 y HsL - 2 + y HsL = s2 1 s2 y HsL + y HsL = + 2 s2 1 Hs2 + 1L y HsL = + 2 s2 1 2 y HsL = + . s2 H s2 + 1L H s2 + 1L
Using partial fractions, we set 1 A B = + 2 2 2 2 s H s + 1L s H s + 1L
Using the cover−up method (first set s=0 and then s2 =−1) we find that A = 1, B = -1. Hence 1 1 2 y HsL = - + s2 Hs2 + 1L Hs2 + 1L
1 1 y HsL = + 2 2 s Hs + 1L y HtL = t + sin HtL.
Solution 4.4 To solve the initial value problem y ’’ + 2 y ’ + y = 3 t e- t , y H0L = 4, y ’ H0L = 2;
we take the Laplace transform of both sides of the differential equation À Hy ’’ HtL + 2 y ’ HtL + y HtLL = À H3 t e- t L
À Hy ’’ HtLL + 2 À Hy ’ HtLL + À Hy HtLL = 3 À Ht e- t L 1 s2 y HsL - s y H0L - y ’ H0L + 2 Hs y HsL - y H0LL + y HsL = 3 Hs + 1L2 1 s2 y HsL - 4 s - 2 + 2 Hs y HsL - 4L + y HsL = 3 Hs + 1L2 1 s2 y HsL + 2 s y HsL + y HsL = 3 + 10 + 4 s Hs + 1L2 1 Hs2 + 2 s + 1L y HsL = 3 + 10 + 4 s Hs + 1L2 1 Hs + 1L2 y HsL = 3 + 10 + 4 s Hs + 1L2 1 10 + 4 s y HsL = 3 + . 4 Hs + 1L Hs + 1L2
Now notice that 10 + 4 s 4s+4+6 4 Hs + 1L + 6 4 6 = = = + 2 2 2 Hs + 1L Hs + 1L Hs + 1L Hs + 1L Hs + 1L2
Hence
1 1 1 y HsL = 3 + 4 + 6 4 Hs + 1L Hs + 1L Hs + 1L2
1 y HtL = ã-t t3 + 4 ã- t + 6 ã- t t. 2
Solution 4.5 To solve the initial value problem y ’’ + 16 y = 32 t,
y H0L = 3, y ’ H0L = -2;
we take the Laplace transform of both sides of the differential equation
À Hy ’’ HtL + y HtLL = À H32 tL
À Hy ’’ HtLL + À Hy HtLL = 32 À HtL
1 s2 y HsL - s y H0L - y ’ H0L + 16 y HsL = 32 s2 1 s2 y HsL - 3 s + 2 + 16 y HsL = 32 s2 32 Hs2 + 16L y HsL = - 2 + 3 s s2 32 2 3s y HsL = - + s2 Hs2 + 16L Hs2 + 16L Hs2 + 16 L 2 2 2 3s y HsL = - - + 2 2 2 2 s Hs + 16 L Hs + 16L Hs + 16 L 2 4 3s y HsL = - + 2 2 2 s Hs + 16L Hs + 16 L y HtL = 2 t - sin H4 tL + 3 cos H4 tL.
32 . Note that we used partial fractions to split the term s2 Hs2 +16L
Solution 4.6 To solve the initial value problem y ’’ - 3 y ’ + 2 y = 4,
y H0L = 1, y ’ H0L = 0;
we take the Laplace transform of both sides of the differential equation
À Hy ’’ - 3 y ’ + 2 yL = À H4L
À Hy ’’ HtLL - 3 À Hy ’ HtLL + 2 À Hy HtLL = À H4L
4 s2 y HsL - s y H0L - y ’ H0L - 3 Hs y HsL - y H0LL + 2 y HsL = s 4 s2 y HsL - s - 0 - 3 Hs y HsL - 1L + 2 y HsL = s 4 Hs2 - 3 s + 2L y HsL = + s - 3 s 4 Hs - 1L Hs - 2L y HsL = + s - 3 s
4 s-3 y HsL = + . s Hs - 1L Hs - 2L Hs - 1L Hs - 2L
Using partial fractions, set 4 A B C = + + s Hs - 1L Hs - 2L s Hs - 1L Hs - 2L
4 = A Hs - 1L Hs - 2L + B s Hs - 2L + C s Hs - 1L.
Using the cover−up method we soon see that A = 2, B = -4, C = 2.
Also using partial fractions, set s-3 D E = + Hs - 1L Hs - 2L Hs - 1L Hs - 2L
s - 3 = D Hs - 2L + E Hs - 1L.
Using the cover−up method we soon see that D = 2, E = -1.
Hence 2 4 2 2 1 y HsL = - + + - s Hs - 1L Hs - 2L Hs - 1L Hs - 2L
2 2 1 y HsL = - + s Hs - 1L Hs - 2L
y HtL = 2 - 2 ãt + ã2 t .
Solution 4.7 To solve the initial value problem y ’’ + 4 y ’ + 4 y = 6 ã-2 t ,
y H0L = -2, y ’ H0L = 8;
we take the Laplace transform of both sides of the differential equation
À Hy ’’ + 4 y ’ + 4 yL = À H6 ã-2 t L
À Hy ’’ HtLL + 4 À Hy ’ HtLL + 4 À Hy HtLL = 6 À Hã-2 t L
6 s2 y HsL - s y H0L - y ’ H0L + 4 Hs y HsL - y H0LL + 4 y HsL = s+2 6 s2 y HsL + 2 s - 8 + 4 Hs y HsL + 2L + 4 y HsL = s+2 6 Hs2 + 4 s + 4L y HsL = - 2 s s+2 6 Hs + 2L2 y HsL = - 2 s s+2
6 2s y HsL = - 3 Hs + 2L Hs + 2L2
6 2 Hs + 2L - 4 y HsL = - 3 Hs + 2L Hs + 2L2
4 6 i 2 y j z - z y HsL = - j z j 2 s + 2 Hs + 2L Hs + 2L3 k {
6 2 4 y HsL = - + 3 s+2 Hs + 2L Hs + 2L2
y HtL = ã-2 t × 3 t2 - ã-2 t × 2 + ã-2 t × 4 t
y HtL = ã-2 t H3 t2 + 4 t - 2L.
Solution 4.8 To solve the initial value problem y ’’ + 4 y ’ + 5 y = ∆ Ht - 4 ΠL,
y H0L = 0, y ’ H0L = 3;
we take the Laplace transform of both sides of the differential equation À Hy ’’ HtL + 4 y ’ HtL + 5 y HtLL = À H∆ Ht - 4 ΠLL
À Hy ’’ HtLL + 4 À Hy ’ HtLL + 5 À Hy HtLL = À H∆ Ht - 4 ΠLL s2 y HsL - s y H0L - y ’ H0L + 4 Hs y HsL - y H0LL + 5 y HsL = ã-4 Π s s2 y HsL - 0 - 3 + 4 Hs y HsL - 0L + 5 y HsL = ã-4 Π s Hs2 + 4 s + 5L y HsL = ã-4 Π s + 3
ã-4 Π s 3 y HsL = + 2 2 s +4s+5 s +4s+5 ã-4 Π s 3 y HsL = + 2 Hs + 2L + 1 Hs + 2L2 + 1
1 1 i y j z y HsL = ã-4 Π s j z j z + 3 × 2 Hs + 2L2 + 1 k Hs + 2L + 1 { y HsL = ã-4 Π s g HsL + 3 × g HsL
where 1 g HsL = = À Hã-2 t sin HtLL Hs + 2L2 + 1
and ã-4 Π s g HsL = À Hã-2 Ht-4 ΠL sin Ht - 4 ΠLL.
Hence y HtL = 9
ã-2 Ht-4 ΠL sin Ht - 4 ΠL + 3 ã-2 t sin HtL, 3 ã-2 t sin HtL,
t > 4 Π, t £ 4 Π.
Plot@UnitStep@t - 4 ΠD Hã-2 Ht-4 ΠL Sin@t - 4 ΠDL + 3 ã-2 t Sin@tD, 8t, 0, 8 Π
