Possible ways of solving second order ODE IVP.

Question

The problem is the following:

y''(t) + 2y'(t) + 5y(t) = H(t-2) - H(t-1) (Where H(t) is the Heaviside function)

y(0)=0, y'(0)=0 (Initial values)

I started working with the Laplace transform and ended up with:

L[y(t)]=( 1/(s^2+2s+5) ) + ....

This is where i am stuck, since the Polynomial above has no real solution i cant break it to two first-degree polynomials.

My questions are whether using Laplace transform is even the right way to solve this problem, and if so, how do you deal with the polynomial issue.

Thanks.

Answer 1

We have $\mathcal L^{-1}(\frac1{s^2+a^2})=\frac{\sin(a t)}a$

So $\mathcal L(\sin(2t)/2)=\frac1{s^2+4}$

And $\frac1{s^2+2s+5}=\frac1{(s+1)^2+2^2}$

So all we need to do is shift $\mathcal L(\sin(2t)/2)$, we can do this by multiplying $f(t)$ by $e^{-t}$.

Thus$$\mathcal L(0.5e^{-t}\sin(2t))=\frac1{s^2+2s+5}\\\implies \mathcal L^{-1}(\frac1{s^2+2s+5})=0.5e^{-t}\sin(2t)$$