Question

Use the Laplace Transform to solve the initial-value problem: `y''+2y'+y=delta(t-2)`; `y(0)=0`, `y'(0)=0`.

Answer

`y''+2y'+y=delta(t-2)`, `y(0)=0,y'(0)=0`.

Take Laplace transform of both sides:

`L(y''+2y'+y)=L(delta(t-2))`.

`L(y'')+2L(y')+L(y)=L(delta(t-2))`.

Using the following formulas: `L(y'')=s^2Y(s)-sy(0)-y'(0)`, `L(y')=sY(s)-y(0)`,

`L(delta(t-2))=e^(-2s)`, where `L(y)=Y(s)`, we can rewrite equation as follows:

`s^2Y(s)-sy(0)-y'(0)+2(sY(s)-y(0))=Y(s)=e^(-2s)`.

Applying initial conditions gives:

`s^2Y(s)-s*0-0+2(sY(s)-0)+Y(s)=e^(-2s)`

Or

`(s^2+2s+1)Y(s)=e^(-2s)`.

`Y(s)=e^(-2s)/(s^2+2s+1)=e^(-2s)/(s+1)^2`

So, since `L^-1(1/(s+1)^2)=te^-t` then

`y(t)=L^-1(Y(s))=L^-1(e^(-2s)/(s+1)^2)=u_2(t)e^((-(t-2)))(t-2)=u_2(t)e^((2-t))(t-2)`.

Answer: `y(t)=u_2(t)e^((2-t))(t-2)`.