Question

# Use Laplace transform to solve the folowing initial value problem y"+2y'+y=4e^{-t} y(0)=2 y'(0)=-1

Laplace transform
Use Laplace transform to solve the folowing initial value problem $$y"+2y'+y=4e^{-t} y(0)=2 y'(0)=-1$$

2021-01-23
Step 1
Given : $$y"+2y'+y=4e^{-t} y(0)=2 y'(0)=-1$$
Applying Laplace Transform
$$L(y")+2L(y')+L(y)=4L(e^(t))$$
$$\Rightarrow s^2Y(s)-sy(0)-y'(0)+2[sY(s)-y(0)]+Y(s)=\frac{4}{(s+1)}$$
$$\Rightarrow (s^2+2s+1)Y(s)-2s+1-4=\frac{4}{(s+1)} [y(0)=2 , y'(0)=-1]$$
$$\Rightarrow (s+1)^2Y(s)=\frac{4}{(s+1)}+2s+3$$
$$\Rightarrow Y(s)=\frac{4}{(s+1)^3}+\frac{(2s+3)}{(s+1)^2}$$
$$\Rightarrow Y(s)=\frac{4}{(s+1)^3}+\frac{2(s+\frac{3}{2}+1-1)}{(s+1)^2}$$
$$\Rightarrow Y(s)=\frac{4}{(s+1)^3}+\frac{2(s+1)}{(s+1)^2}+\frac{1}{(s+1)^2}$$
$$Y(s)=\frac{4}{(s+1)^3}+\frac{2(s+1)}{(s+1)^2}+\frac{1}{(s+1)^2}$$
Step 2
Now taking inverse Laplace Transform,
$$L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{4}{(s+1)^3}\right\}+2L^{-1}\left\{\frac{2(s+1)}{(s+1)^2}\right\}+L^{-1}\left\{\frac{1}{(s+1)^2}\right\}$$
Using the identity $$L^{-1}\left\{\frac{1}{(s+a)^n}\right\}=\frac{t^{n-1}e^{-at}}{(n-1)!}$$ , we get
$$y(t)=2t^2e^{-t}+2e^{-t}+te^{-t}$$