Given the IVP: $$y''+2y'+5y=50t-100$$ $$y(2) = -4$$ $$y'(2)=14$$
Solve the IVP.
I fairly certain that this type of problem requires using the second shifting theorem. First I apply the theorem to get:
$$\widetilde{t} = t-2$$ $$y'(t) = \widetilde{y}(\widetilde{t})$$
But I'm not sure how to proceed, or if this is even relevant. I'm not sure that I understand how to apply the second shifting theorem to this problem at all.
The reason for the shifting is so that the initial values will shift to $y(0)$ and $y'(0)$. Since the Laplace transform of the $y'$ and $y''$ are dependent on $y(0)$ and $y'(0)$, this makes solving the problem easier.
The shift gives the equivalent IVP $$ \overline{y}'' + 2\overline{y}' + \overline{y} = 50\overline{t}, \quad \overline{y}(0) = -4, \quad \overline{y}'(0) = 14$$
The next step is to find $\overline{y}(\overline{t})$ by taking the Laplace transform of the equation, solve for $Y(s)$ then inverse transform back.
The solution to the original equation can be found by shifting back $$ y(t) = \overline{y}(\overline{t}+2)$$