I need to solve the following initial value problem using Laplace transform method:
$$y''+2y'+y=t+\delta(t) \quad (t>-1);\quad y(-1)=0,\quad y'(-1)=1$$
where the function on the right side is Dirac Delta function. But when I take Laplace transform of the both side, there will be $y(0)$ term which is not given in the problem. So how do I do it?
Best Answer
Let $t=x-1$ and we have initial value problem $$y''+2y'+y=(x-1)+\delta(x-1)~~~;~~~x>0,\quad y(0)=0,\quad y'(0)=1$$ so we have $${\cal L}(y)(s+1)^2=\dfrac{1}{s^2}-\dfrac1s+1+e^{-s}$$ after finding the solution $$y(x)=(x+3)e^{-x}+(x-3)+e^{1-x}(x-1){\bf H}_1(x)$$ where ${\bf H}$ is Heaviside Step Function. Feel free to get back with substitution $x-1=t$.