solns4.nb 1 Chapter 4 (Laplace transforms): Solutions

1 Chapter 4 ( laplace transforms): Solutions (The table of laplace transforms is used throughout.) Solution (a) HsinH4 tLcos H2 tLL= ikjj1 2sinH4 tLy{zz=1 2 HsinH4 tLL=1 24 s2+16=2 s2+16. Solution (b) Hcosh2 HtL L= ikjjjjikjj1 2 H t- -tLy{zz2y{zzzz= ikjjjj 2 t 4+1 2+ -2 t 4y{zzzz=1 4 1 s-2+1 2 1 s+1 4 1 s+2=s2-2 sHs2-4L. Solution (c) Hcos Ha tLsinh Ha tLL= ikjj1 2 H a t- -a tL cos Ha tLy{zz=1 2 H a tcos Ha tL- -a tcos Ha tLL=1 2ikjjjjs-a Hs-aL2+a2-s+a Hs+aL2+a2y{zzzz=a s2-2 a3 s4+4 a4. Solution (d) Ht2 -3 tL= H -3 tt2L=2 Hs+ Solution (a)Use partial fractions, set s Hs+3L Hs+5L=A Hs+3L+B Hs+5L s=A Hs+5L+B Hs+3 LUsing the cover up method for example we see thatA= -3 2, B=5 Hs+3L Hs+5L= -3 2 1 Hs+3L+5 2 1 Hs+5L= ikjj-3 2 -3 ty{zz+ ikjj5 2 -5 ty{ so the inverseLaplacetransform -1 ikjjjs Hs+3L Hs+5Ly{zzz=-3 2 -3 t+5 2 -5 t.}}}}}}}}}

2 Solution (b)Use partial fractions, set 1 sHs2+k2L=A s+B s+C s2+k2 1=A Hs2+k2L+HB s+CLs 1=A s2+A k2+B s2+C s 1=HA+BLs2+C s+A k2 Equating coefficients ofs2: A+B=0s1: C=0s0: A=1 k2 Using the first equation we see that B= -1 k2, and hence1 sHs2+k2L=1 k2 1 s-1 k2 s Hs2+k2L=1 k2 ikjjj1 s-s Hs2+k2Ly{zzz=1 k2 H1-cos Hk so the inverseLaplacetransform -1 ikjjj1 sHs2+k2Ly{zzz=1 k2 H1-cos Hk tLL. Solution (c)To find the inverse laplace transform of 1 Hs+3L2, note that from the table ofLaplace transforms, the laplace transform of t is 1 s2, and so if we apply theshift theorem, the laplace transform of -3 tt must be 1 Hs+3L2.}}

3 Hence -1 ikjjjj1 Hs+3L2y{zzzz=t -3 t. Solution 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 Hy HtL+y HtLL= HtL Hy HtLL+ Hy HtLL= HtL s2y HsL-s y H0L-y H0L+y HsL=1 s2 s2y HsL-2+y HsL=1 s2 s2y HsL+y HsL=1 s2+2 Hs2+1Ly HsL=1 s2+2 y HsL=1 s2 Hs2+1L+2 Hs2+ partial fractions, we set1 s2 Hs2+1L=A s2+B Hs2+1 LUsing the cover up method (first set s=0 and then s2= 1) we find that A=1,B= -1. Hencey HsL=1 s2-1 Hs2+1L+2 Hs2+1L y HsL=1 s2+1 Hs2+1L y HtL=t+ Solution solve the initial value problem y +2 y +y=3 t e-t, y H0L=4, y H0L=2.}

4 We take the laplace transform of both sides of the differential equation Hy HtL+2 y HtL+y HtLL= H3 t e-tL Hy HtLL+2 Hy HtLL+ Hy HtLL=3 Ht e-tL s2y HsL-s y H0L-y H0L+2 Hs y HsL-y H0LL+y HsL=3 1 Hs+1L2 s2y HsL-4 s-2+2 Hs y HsL-4L+y HsL=3 1 Hs+1L2 s2y HsL+2 s y HsL+y HsL=3 1 Hs+1L2+10+4 s Hs2+2 s+1Ly HsL=3 1 Hs+1L2+10+4 s Hs+1L2y HsL=3 1 Hs+1L2+10+4 s y HsL=3 1 Hs+1L4+10+4 s Hs+ notice that10+4 s Hs+1L2=4 s+4+6 Hs+1L2=4Hs+1L+6 Hs+1L2=4 Hs+1L+6 Hs+1L2 Hence y HsL=3 1 Hs+1L4+41 Hs+1L+61 Hs+1L2 y HtL=1 2 -tt3+4 -t+6 -t t.

5 Solution 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 Hy HtL+y HtLL= H32 tL Hy HtLL+ Hy HtLL=32 s2y HsL-s y H0L-y H0L+16 y HsL=321 s2 s2y HsL-3 s+2+16 y HsL=321 s2 Hs2+16L y HsL=32 s2-2+3 s y HsL=32 s2 Hs2+16L-2 Hs2+16L+3 s Hs2+16L y HsL=2 s2-2 Hs2+16L-2 Hs2+16L+3 s Hs2+16L y HsL=2 s2-4 Hs2+16L+3 s Hs2+16L y HtL=2 t-sinH4 tL+3 cosH4 that we used partial fractions to split the term 32 s2 Hs2+16L.

6 Solution 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 Hy -3 y +2 yL= H4L Hy HtLL-3 Hy HtLL+2 Hy HtLL= H4L s2y HsL-s y H0L-y H0L-3Hs y HsL-y H0LL+2 y HsL=4 s s2y HsL-s-0-3Hs y HsL-1L+2 y HsL=4 s Hs2-3 s+2L y HsL=4 s+s-3 Hs-1L Hs-2L y HsL=4 s+s-3 y HsL=4 sHs-1L Hs-2L+s-3 Hs-1L partial fractions, set4 sHs-1L Hs-2L=A s+B Hs-1L+C Hs-2L 4=AHs-1L Hs-2L+B sHs-2L+C the cover up method we soon see that A=2, B= -4, C= using partial fractions, sets-3 Hs-1L Hs-2L=D Hs-1L+E Hs-2L s-3=DHs-2L+ the cover up method we soon see that D=2, E= HsL=2 s-4 Hs-1L+2 Hs-2L+2 Hs-1L-1 Hs-2L y HsL=2 s-2 Hs-1L+1 Hs-2L y HtL=2-2 t+ 2 t.

7 Solution solve the initial value problem y +4 y +4 y=6 -2 t, y H0L= -2, y H0L=8;we take the laplace transform of both sides of the differential equation Hy +4 y +4 yL= H6 -2 tL Hy HtLL+4 Hy HtLL+4 Hy HtLL=6 H -2 tL s2y HsL-s y H0L-y H0L+4Hs y HsL-y H0LL+4 y HsL=6 s+2 s2y HsL+2 s-8+4Hs y HsL+2L+4 y HsL=6 s+2 Hs2+4 s+4L y HsL=6 s+2-2 s Hs+2L2 y HsL=6 s+2-2 s y HsL=6 Hs+2L3-2 s Hs+2L2 y HsL=6 Hs+2L3-2 Hs+2L-4 Hs+2L2 y HsL=6 Hs+2L3-ikjjjj2 s+2-4 Hs+2L2y{zzzz y HsL=6 Hs+2L3-2 s+2+4 Hs+2L2 y HtL= -2 t 3 t2- -2 t 2+ -2 t 4 y HtL= -2 t H3 t2+4 t-2L.}

8 Solution solve the initial value problem y +4 y +5 y= d Ht-4pL, y H0L=0, y H0L=3;we take the laplace transform of both sides of the differential equation Hy HtL+4 y HtL+5 y HtLL= Hd Ht-4pLL Hy HtLL+4 Hy HtLL+5 Hy HtLL= Hd Ht-4pLL s2y HsL-s y H0L-y H0L+4Hs y HsL-y H0LL+5 y HsL= -4ps s2y HsL-0-3+4Hs y HsL-0L+5 y HsL= -4ps Hs2+4 s+5L y HsL= -4ps+3 y HsL= -4ps s2+4 s+5+3 s2+4 s+5 y HsL= -4ps Hs+2L2+1+3 Hs+2L2+1 y HsL= -4psikjjjj1 Hs+2L2+1y{zzzz+3 1 Hs+2L2+1 y HsL= -4psg HsL+3 g HsLwhereg HsL=1 Hs+2L2+1= H -2 tsin HtLLand -4psg HsL= H -2Ht-4pLsin HtL=9 -2Ht-4pLsin Ht-4pL+3 -2 tsin HtL, t>4 p,3 -2 tsin HtL.}

9 T 4 @UnitStep@t-4pD H -2Ht-4pLSin@t-4pDL+3 -2 tSin@tD,8t, 0, 8 p<, PlotRange Graphics