# With the aid of Laplace Transform, solve the Initial Value Problem {y}text{}{left({t}right)}-{y}'{left({t}right)}={e}^{t} cos{{left({t}right)}}+ sin{{left({t}right)}},{y}{left({0}right)}={0},{y}'{left({0}right)}={0}

With the aid of Laplace Transform, solve the Initial Value Problem
$y\left(t\right)-{y}^{\prime }\left(t\right)={e}^{t}\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right),y\left(0\right)=0,{y}^{\prime }\left(0\right)=0$
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Step 1
The Laplace Transform of a second order linear differential equation ,

is given by,
$Y\left(s\right)=\frac{\left(s+a\right)A+B}{{s}^{2}+as+b}+\frac{F\left(s\right)}{{s}^{2}+as+b}$
Given,
$y\left(t\right)-{y}^{\prime }\left(t\right)={e}^{t}\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right),y\left(0\right)=0,{y}^{\prime }\left(0\right)=0$
Step 2
Applying Laplace Transform in the overall equation (1),
$y\left(t\right)-{y}^{\prime }\left(t\right)={e}^{t}\mathrm{cos}\left(t\right)+\mathrm{sin}\left(t\right)$
$⇒L\left[y\left(t\right),s\right]-L\left[{y}^{\prime }\left(t\right),s\right]=L\left[{e}^{t}\mathrm{cos}\left(t\right),s\right]+L\left[\mathrm{sin}\left(t\right),s\right]$
$⇒\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]-\left[sY\left(s\right)-y\left(0\right)\right]=L{\left[\mathrm{cos}\left(t\right),s\right]}_{s}\to s-1+\frac{a}{{s}^{2}+{1}^{2}}$
$⇒{s}^{2}Y-sY\left(s\right)={\left[\frac{s}{{s}^{2}+1}\right]}_{s}\to s-1+\frac{1}{{s}^{2}+1}$
$⇒\left({s}^{2}-s\right)Y\left(s\right)=\left[\frac{s-1}{{\left(s-1\right)}^{2}+1}\right]+\frac{1}{{s}^{2}+1}$
$⇒s\left(s-1\right)Y\left(s\right)=\left[\frac{s-1}{{\left(s-1\right)}^{2}+1}\right]+\frac{1}{{s}^{2}+1}$
$⇒Y\left(s\right)=\frac{1}{s\left[{\left(s-1\right)}^{2}+1\right]}+\frac{1}{s\left(s-1\right)\left({s}^{2}+1\right)}$
Step 3
The answer is :$Y\left(s\right)=\frac{1}{s\left[{\left(s-1\right)}^{2}+1\right]}+\frac{1}{s\left(s-1\right)\left({s}^{2}+1\right)}$