# Find the solution of the initial value problem given below by Laplace transform y'-y=t e^t sin t y(0)=0

Laplace transform
Find the solution of the initial value problem given below by Laplace transform
$$y'-y=t e^t \sin t$$
$$y(0)=0$$

2021-03-07
Step 1
The answer is given by using properties of Laplace transform.
Step 2
$$y'-y=t e^t \sin(t)$$
$$y(0)=0$$
Apply the laplace transform
$$L\left\{y'\right\}-L\left\{y\right\}=L\left\{t e^t \sin(t)\right\}$$
$$\Rightarrow (sY(s)-y(0))-Y(s)=\frac{2(s-1)}{(s^2-2s+2)^2}$$
$$\Rightarrow sY(s)-Y(s)=\frac{2(s-1)}{(s^2-2s+2)^2}$$
$$\Rightarrow (s-1)Y(s)=\frac{2(s-1)}{(s^2-2s+2)^2}$$
$$\Rightarrow Y(s)=\frac{2}{(s^2-2s+2)^2}$$
Now apply inverse laplace transform
$$\Rightarrow L^{-1}\left\{Y(s)\right\}=L^{-1}\left\{\frac{2}{(s^2-2s+2)^2}\right\}$$
$$\Rightarrow y(t)=L^{-1}\left\{\frac{2}{((s-1)^2+1)^2}\right\}$$
$$\Rightarrow y(t)=e^t \sin(t)-t e^t \cos(t)$$
$$\Rightarrow y(t)=e^t(\sin(t)-t\cos(t))$$