Question

# By using Laplace transforms, solve the following differential equations subject to the given initial conditions. y"-4y'=-4te^{2t}, y_0=0, y'_0 =1 y"+2y'+10y=-6e^{-t}sin3t, y_0=0, y'_0=1

Laplace transform
By using Laplace transforms, solve the following differential equations subject to the given initial conditions.
$$y"-4y'=-4te^{2t}, y_0=0, y'_0 =1$$
$$y"+2y'+10y=-6e^{-t}\sin3t, y_0=0, y'_0=1$$

2021-02-16
Step 1 The given differential equation is,
$$y"-4y'=-4te^{2t}$$
where $$y_0=0, y'_0 =1$$
Take the Laplace Transform on both sides,
$$L(y''-4y')=L(-4te^{2t})$$
$$L(y'')-4L(y')=-4L(te^{2t})$$
Consider the formulae,
$$L(y'')=s^2L(y)-sy_0-y_0'$$
$$L(y')=sL(y)-y_0$$
$$L(te^{2t})=\frac{1}{(s-2)^2}$$
Step 2
Substitute the values,
$$L(y'')-4L(y')=-4L(te^{2t})$$
$$s^2L(y)-sy_0-y_0'-4[sL(y)-y0]=-4 \cdot \frac{1}{(s-2)^2}$$
$$s^2L(y)-s \cdot (0)-(1)-4sL(y)=\frac{(-4)}{(s-2)^2}$$
$$L(y) \cdot [s^2-4s]=1+\frac{(-4)}{(s-2)^2}$$
$$L(y)=\frac{s^2-4s+4-4}{(s-2)^2(s^2-4s)}$$
$$L(y)=\frac{1}{(s-2)^2}$$
Step 3
Take the Inverse Laplace Transform both sides,
$$L^{-1}[L(y)]=L^{-1}\left[\frac{1}{(s-2)^2}\right]$$
$$y=L^{-1}\left[\frac{1}{(s-2)^2}\right]$$
$$y=te^{2t}$$
Thus, the solution of the given integral is $$y=te^{2t}$$ ii)
The given differential equation is,
$$y''+24y'+10y=-6e^{-t}\sin{3t}$$
$$\text{where } y_0=0 , y_0'=1$$
Step 4
Take the Laplace Transform on both sides,
$$L(y''+24y'+10y)=L[-6e^{-t}\sin(3t)]$$
$$L(y'')+24L(y')+10L(y)=-6L[e^{-t}\sin(3t)]$$
Use the same formulae as mentioned above,
$$s^2L(y)-0-1+2[sL(y)-0]+10L(y)=(-6) \cdot \frac{3}{(s+1)^2+9}$$
$$s^2L(y)-1+2sL(y)+10L(y)=\frac{-18}{s^2+2s+1+9}$$
$$[s^2+2s+10]L(y)=1-\frac{18}{s^2+2s+10}$$
$$L(y)=\frac{s^2+2s+10-18}{(s^2+2s+10)^2}$$
$$L(y)=\frac{s^2+2s-8}{(s^2+2s+10)^2}$$
Step 5
Take the Inverse Laplace Transform both sides,
$$L^{-1}[L(y)]=L^{-1}\left[\frac{s^2+2s-8}{(s^2+2s+10)^2}\right]$$
$$y=te^{-t}\cos(3t)$$
Thus, the solution of the given differential equation is $$y=te^{-t}\cos(3t)$$