# Find the Laplace transform Y(s), of the solution of the IVP y"+3y'+2y=cos(2t) y(0)=0 y'(0)=1 Do not solve the IVP

Find the Laplace transform Y(s), of the solution of the IVP
$y"+3{y}^{\prime }+2y=\mathrm{cos}\left(2t\right)$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=1$
Do not solve the IVP
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Step 1
The given initial value problem is ${y}^{″}+3{y}^{\prime }+2y=\mathrm{cos}\left(2t\right),y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$
Find the Laplace transform Y(s) of the solution of the IVP as follows.
Step 2Consider the differential equation ${y}^{″}+3{y}^{\prime }+2y=\mathrm{cos}\left(2t\right)$
Take Laplace transform on both sides and obtain,
$L\left\{y"+3{y}^{\prime }+2y\right\}=L\left\{\mathrm{cos}\left(2t\right)\right\}$
$L\left\{y"\right\}+3L\left\{{y}^{\prime }\right\}+2L\left\{y\right\}=L\left\{\mathrm{cos}\left(2t\right)\right\}$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+3\left(sY\left(s\right)-y\left(0\right)\right)+2\left(Y\left(s\right)\right)=\frac{s}{\left({s}^{2}+4\right)}$
${s}^{2}Y\left(s\right)-sy\left(0\right)-1+3\left(sY\left(s\right)-y\left(0\right)\right)+2\left(Y\left(s\right)\right)=\frac{s}{\left({s}^{2}+4\right)}$
${s}^{2}Y\left(s\right)-1+3sY\left(s\right)+2Y\left(s\right)=\frac{s}{\left({s}^{2}+4\right)}$
$Y\left(s\right)\left({s}^{2}+3s+2\right)-1=\frac{s}{\left({s}^{2}+4\right)}$
$Y\left(s\right)\left({s}^{2}+3s+2\right)=\frac{s}{\left({s}^{2}+4\right)+1}$
$Y\left(s\right)=\frac{s+\left({s}^{2}+4\right)}{\left({s}^{2}+4\right)\left({s}^{2}+3s+2\right)}$
$Y\left(s\right)=\frac{\left({s}^{2}+s+4\right)}{\left({s}^{2}+4\right)\left({s}^{2}+3s+2\right)}$
Step 3
Therefore, the Laplace transform Y(s) of the solution of the IVP is,$Y\left(s\right)=\frac{\left({s}^{2}+s+4\right)}{\left({s}^{2}+4\right)\left({s}^{2}+3s+2\right)}$