find the solution to the initial value problem using the laplace Transform. xy"+(3-x)y'-2y=x-1, y(0)=0 , y'(0)=\frac{1}{3}

Kyran Hudson 2021-09-05 Answered

find the solution to the initial value problem using the laplace Transform.
xy"+(3x)y2y=x1
y(0)=0,y(0)=13

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Expert Answer

Aubree Mcintyre
Answered 2021-09-06 Author has 73 answers

Step 1
Given IVP is
xy(3x)y2y=x1 with (1)
y(0)=0,y(0)=12
We know that
L[y]=s2L[y]sy(0)y(0)
L[y]=sL[y]y(0)
L[fg]=L[f]L[g]
L[xn]=sn+1n!  and  L[f+g]=L[f]+L[g]
Taking Laplace transform of eqn (1)
L[xy]+L[(3x)y]L[2y]=L[x1]
L[x]L[y]+L[3x]L[y]2L[y]=L[x1]
s21![s2L[y]sy(0)y(0)]+[3L[1]L[x]]×L[y]2s=L[x]L[1]
s4L[y]s30s2×13+[3ss2][sL[y]y(o)]2s=s2s
s4L[y]s23+3s2L[y]s3L[y]=2s+s2s
(s4s3+3s2)L[y]=s+s2+s23

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