Question

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

Answer 1

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
\(\Rightarrow L^{-1}\left\{Y(s)\right\}=L^{-1}\left\{\frac{2}{(s^2-2s+2)^2}\right\}\)
\(\Rightarrow y(t)=L^{-1}\left\{\frac{2}{((s-1)^2+1)^2}\right\}\)
\(\Rightarrow y(t)=e^t \sin(t)-t e^t \cos(t)\)
\(\Rightarrow y(t)=e^t(\sin(t)-t\cos(t))\)