Consider the following differential equation, where ω and µ are non-zero positive real numbers, and $f$ a causal function:
$$x''(t)+\omega^2x(t)=f(t) \quad\text{where}\quad f(t)=e^{-\mu t}$$
The first question is: Determine the solution of the differential equation $x(t), t > 0$ using the transform when $x(0) = x'(0) = 0$. Justify all the steps of your reasoning and calculations.
Here is my initial idea, but I am stuck at the end.
I know that
$$TL[x]=X(p)\\TL[x’]=p.X(p)-x(0)=p.X(p)-0=p.X(p)\\TL[x’’]=p^2.X(p)-p.x(0)-x’(0)=p^2.X(p)-p.0-0=p^2.X(p)$$
With
$$e^{-\mu t}=\frac1{p+\mu}$$
So
$$p^2.X(t)+w^2.X(p)=\frac1{p+\mu}\\X(t)[p^2+w^2]=\frac1{p+\mu}\\X(t)=\frac1{p+\mu}\cdot\frac1{p^2+w^2}=…\\ \\(-1)=i^2\\p^2-i^2.w^2$$ I would like your opinion.