I am trying to solve the following equation: $$y'' + y' + y = 1 + e^{-t}$$
After applying the Laplace transform and using partial fractions I have terms of the form:
$$\frac{1}{s^2 + s + 1}$$ and $$\frac{s}{s^2+s+1}$$
which I don't know how to reverse transform. Could you help me?
You didn't give the initial conditions...a bit hard to check your solution
You have that $$\frac{s}{s^2 + s + 1}=\frac{s}{(s^2 + s + 1/4 +3/4)}=\frac{s}{(s+1/2)^2 + 3/4}$$ $$\frac{s}{(s+1/2)^2 + 3/4}=\frac{s+1/2}{(s+1/2)^2 + 3/4}-\frac{1/2}{(s+1/2)^2 + 3/4}$$ And look at the sine and cosine function in the table... $$\mathcal{L}({e^{at}\cos(\beta t)})=\frac {s-a}{(s-a)^2+\beta ^2 }$$ $$\mathcal{L}({e^{at}\sin(\beta t)})=\frac {\beta}{(s-a)^2+\beta ^2 }$$