Working on a question and was wondering if this is right.
Find the Laplace transform:
$$x'' + 2x' + x = f(t)$$ $$x(0)=x'(0)=0$$ $$f(t) = \begin{cases}2 ,& 0 \leq t \leq 2 \\t ,& t \ge 2\end{cases}$$
Heaviside function: $$2-tu(t-2)$$
The Laplace transform of it is : $$\frac{2}{s} -\frac{1}{s^3}e^{-2s}$$
Then, let $$X = \mathcal{L}{x(t)}$$ $$\mathcal{L}x'' = s^2X - sx(0) - x'(0) = {s^2}X$$ $$\mathcal{L}x' = 2(sX-x(0))= 2sX$$
The transformed equation would be: $$s^2X +2sX + X = \frac{2}{s} - \frac{e^2s}{s^3} => X = \frac{2}{s(s+1)^2} - \frac{e^-2s}{s^3(s+1)^2}$$
Now, to solve for x(t), I have to use partial fraction right? Or is there a faster way!
Also, I have a question regarding the $$\leq and < $$ signs. If I had $$0 \leq t < 2$$ instead, what impact does it have on the equation.
Would someone want to guide me through the convolution? I'm not exactly sure where to start.