Find the solutions for the given linear systems of differential equations using Laplace Transforms. w'-2y'+3w=0: y"+w=2sin x y(0)=w(0)=2 , y’(0)=-1

Clifland

Clifland

Answered question

2021-01-31

Find the solutions for the given linear systems of differential equations using Laplace Transforms.
w2y+3w=0:y"+w=2sinx
y(0)=w(0)=2,y(0)=1

Answer & Explanation

ensojadasH

ensojadasH

Skilled2021-02-01Added 100 answers

Step 1
Consider the given system,
w2y+3w=0:y"+w=2sinx
y(0)=w(0)=2,y(0)=1
Applying Laplace transform on both sides,
L{y}+L{w}=2L{sinx}
s2y(s)sy(0)y(0)+w(s)=2s2+1
s2y(s)2s+1+w(s)=2s2+1
s2y(s)+w(s)=2s2+1+2s1(1)
And, 
Lw2Ly+3Lw=0
sw(s)w(0)2[sy(s)y(0)]+3w(s)=0
(s+3)w(s)2sy(s)=2(2)
Step 2
Multiply the equation (1) with (s+3) and subtract
(s2(s+3)+2s)y(s)=2s3+2s2+6s+2(s+3)(s2+1)
(s2(s+3)+2s)y(s)=2s2+5s1+2(s+3)(s2+1)
y(s)=2s2+5s1s3+3s2+2s+2(s+3)s2+1(s3+3s2+2s)
Resolve into partial fractions,
y(s)=65ss2+185s2+1+52s+2s+11310(s+2)
y(s)=6s5(s2+1)85(s2+1)+52s+2s+11310(s+2)
Apply inverse Laplace transform on both sides,
L1{y(s)}=65L(1){s(s2+1)}85L1{1(s2+1)}+52L1{1s}+2L1{1s+1}1310L1{1(s+2)}
y(x)=65cos(x)85sin(x)+52(1)+2ex1310e2x
Step 3
Multiply equation (1) by 2 and equation by 5 we get, and then adding.

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