 # Find the solution of the initial value problem given below by Laplace transform y'-y=t e^t sin t y(0)=0 generals336 2021-03-06 Answered
Find the solution of the initial value problem given below by Laplace transform
${y}^{\prime }-y=t{e}^{t}\mathrm{sin}t$
$y\left(0\right)=0$
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Step 1
The answer is given by using properties of Laplace transform.
Step 2
${y}^{\prime }-y=t{e}^{t}\mathrm{sin}\left(t\right)$
$y\left(0\right)=0$
Apply the laplace transform
$L\left\{{y}^{\prime }\right\}-L\left\{y\right\}=L\left\{t{e}^{t}\mathrm{sin}\left(t\right)\right\}$
$⇒\left(sY\left(s\right)-y\left(0\right)\right)-Y\left(s\right)=\frac{2\left(s-1\right)}{\left({s}^{2}-2s+2{\right)}^{2}}$
$⇒sY\left(s\right)-Y\left(s\right)=\frac{2\left(s-1\right)}{\left({s}^{2}-2s+2{\right)}^{2}}$
$⇒\left(s-1\right)Y\left(s\right)=\frac{2\left(s-1\right)}{\left({s}^{2}-2s+2{\right)}^{2}}$
$⇒Y\left(s\right)=\frac{2}{\left({s}^{2}-2s+2{\right)}^{2}}$
Now apply inverse laplace transform
$⇒{L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{\frac{2}{\left({s}^{2}-2s+2{\right)}^{2}}\right\}$
$⇒y\left(t\right)={L}^{-1}\left\{\frac{2}{\left(\left(s-1{\right)}^{2}+1{\right)}^{2}}\right\}$
$⇒y\left(t\right)={e}^{t}\mathrm{sin}\left(t\right)-t{e}^{t}\mathrm{cos}\left(t\right)$
$⇒y\left(t\right)={e}^{t}\left(\mathrm{sin}\left(t\right)-t\mathrm{cos}\left(t\right)\right)$