Find the solution of the Differentional equation by using Laplace Transformation2y'-3y=e^{2t}, y(0)=1y"+y=t, y(0)=0 and y'(0)=2

Kaycee Roche 2020-11-08 Answered

Find the solution of the Differentional equation by using Laplace Transformation
2y3y=e2t,y(0)=1
y"+y=t,y(0)=0 and y(0)=2

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Expert Answer

funblogC
Answered 2020-11-09 Author has 91 answers

Step 1
1) Given equation is
2y3y=e2t,y(0)=1
Taking Laplace transform , we get
2L(y)3L(y)=L(e2t)
sY(s)y(0)3Y(s)=1s2
(s+3)Y(s)1=1(s2)[y(0)=1]
Y(s)=1(s+3)(s2)+1(s+3)
Step 2
Using partial fraction decomposition,
1(s+3)(s2)=15(s2)15(s+3)
So,
Y(s)=15(s2)15(s+3)+1(s+3)
Y(s)=15(s2)+45(s+3)
On taking inverse Laplace Transform, we get
y(t)=15e2t+45e3t[L1{1(sa)}=eat]
Step 3
y"+y=t,y(0)=0,y(0)=2
Taking Laplace transform , we get L(y")+L(y)=L(t)
s2Y(s)sy(0)y(0)+Y(s)=1(s2)[L(tn)=(n!)sn+1]
(s2+1)Y(s)2=1(s2)[y(0)=0,y(0)=2]
Y(s)=1(s2+1)s2+2(s2+1)(1)
Using partial fraction decomposition,
1(s2+1)s2=1s21(s2+1)
So, from (1), we get
Y(s)=1s21(s2+1)+2(s2+1)
Y(s)=1s2+1(s2+1)
On taking inverse Laplace Transform, we get
y(t)=t+sin(t)[L1{1sn+1}=tn(n!),L1{a(s2+a2)}=sin(at)]

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