# 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

Find the solutions for the given linear systems of differential equations using Laplace Transforms.
$w-2y+3w=0:y"+w=2\mathrm{sin}x$
$y\left(0\right)=w\left(0\right)=2,{y}^{\prime }\left(0\right)=-1$
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Step 1
Consider the given system,
${w}^{\prime }-2{y}^{\prime }+3w=0:y"+w=2\mathrm{sin}x$
$y\left(0\right)=w\left(0\right)=2,{y}^{\prime }\left(0\right)=-1$
Applying Laplace transform on both sides,
$L\left\{{y}^{″}\right\}+L\left\{w\right\}=2L\left\{\mathrm{sin}x\right\}$
${s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+w\left(s\right)=\frac{2}{{s}^{2}+1}$
${s}^{2}y\left(s\right)-2s+1+w\left(s\right)=\frac{2}{{s}^{2}+1}$
${s}^{2}y\left(s\right)+w\left(s\right)=\frac{2}{{s}^{2}+1}+2s-1\dots \left(1\right)$

$L{w}^{\prime }-2L{y}^{\prime }+3Lw=0$
$sw\left(s\right)-w\left(0\right)-2\left[sy\left(s\right)-y\left(0\right)\right]+3w\left(s\right)=0$
$\left(s+3\right)w\left(s\right)-2sy\left(s\right)=-2\dots \left(2\right)$
Step 2
Multiply the equation (1) with (s+3) and subtract
$\left(s2\left(s+3\right)+2s\right)y\left(s\right)=2-s-3+2{s}^{2}+6s+\frac{2\left(s+3\right)}{\left({s}^{2}+1\right)}$
$\left(s2\left(s+3\right)+2s\right)y\left(s\right)=2{s}^{2}+5s-1+\frac{2\left(s+3\right)}{\left({s}^{2}+1\right)}$
$y\left(s\right)=\frac{2{s}^{2}+5s-1}{{s}^{3}+3{s}^{2}+2s}+\frac{2\left(s+3\right)}{{s}^{2}+1\left({s}^{3}+3{s}^{2}+2s\right)}$
Resolve into partial fractions,
$y\left(s\right)=\frac{\frac{-6}{5s}}{{s}^{2}+1}-\frac{\frac{8}{5}}{{s}^{2}+1}+\frac{5}{2s}+\frac{2}{s+1}-\frac{13}{10\left(s+2\right)}$
$y\left(s\right)=\frac{-6s}{5\left({s}^{2}+1\right)}-\frac{8}{5\left({s}^{2}+1\right)}+\frac{5}{2s}+\frac{2}{s+1}-\frac{13}{10\left(s+2\right)}$
Apply inverse Laplace transform on both sides,
${L}^{-1}\left\{y\left(s\right)\right\}=-\frac{6}{5}{L}^{\left(}-1\right)\left\{\frac{s}{\left({s}^{2}+1\right)}\right\}-\frac{8}{5}{L}^{-1}\left\{\frac{1}{\left({s}^{2}+1\right)}\right\}+\frac{5}{2}{L}^{-1}\left\{\frac{1}{s}\right\}+2{L}^{-1}\left\{\frac{1}{s+1}\right\}-\frac{13}{10}{L}^{-1}\left\{\frac{1}{\left(s+2\right)}\right\}$
$y\left(x\right)=-\frac{6}{5}\mathrm{cos}\left(x\right)-\frac{8}{5}\mathrm{sin}\left(x\right)+\frac{5}{2}\left(1\right)+2{e}^{-x}-\frac{13}{10}{e}^{-2x}$
Step 3
Multiply equation (1) by 2 and equation by 5 we get, and then adding.