# 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}

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

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Step 1
Given IVP is
$xy\left(3-x\right){y}^{\prime }-2y=x-1$ with (1)
$y\left(0\right)=0,{y}^{\prime }\left(0\right)=\frac{1}{2}$
We know that
$L\left[y\right]={s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left[{y}^{\prime }\right]=sL\left[y\right]-y\left(0\right)$
$L\left[f\cdot g\right]=L\left[f\right]\cdot L\left[g\right]$

Taking Laplace transform of eqn (1)
$L\left[xy\right]+L\left[\left(3-x\right){y}^{\prime }\right]-L\left[2y\right]=L\left[x-1\right]$
$\therefore L\left[x\right]\cdot L\left[y\right]+L\left[3-x\right]\cdot L\left[{y}^{\prime }\right]-2L\left[y\right]=L\left[x-1\right]$
$\therefore \frac{{s}^{2}}{1!}\left[{s}^{2}L\left[y\right]-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+\left[3L\left[1\right]-L\left[x\right]\right]×L\left[{y}^{\prime }\right]-2s=L\left[x\right]-L\left[1\right]$
$\therefore {s}^{4}L\left[y\right]-{s}^{3}\cdot 0-{s}^{2}×\frac{1}{3}+\left[3s-{s}^{2}\right]\left[sL\left[y\right]-y\left(o\right)\right]-2s={s}^{2}-s$
$\therefore {s}^{4}L\left[y\right]-\frac{{s}^{2}}{3}+3{s}^{2}L\left[y\right]-{s}^{3}L\left[y\right]=2s+{s}^{2}-s$
$\therefore \left({s}^{4}-{s}^{3}+3{s}^{2}\right)L\left[y\right]=s+{s}^{2}+\frac{{s}^{2}}{3}$

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