# Thread: Solve y'' - y = -2t - 2δ_3(t), y(0) = 1, y'(0) = -1

1. ## Solve y'' - y = -2t - 2δ_3(t), y(0) = 1, y'(0) = -1

The question is attached. Could someone explain to me how to find the solution to y(t) for both conditions for t?

Any help would be greatly appreciated!
Thanks in advance!

2. ## Re: Solve y'' - y = -2t - 2δ_3(t), y(0) = 1, y'(0) = -1

Originally Posted by s3a
The question is attached. Could someone explain to me how to find the solution to y(t) for both conditions for t?

Any help would be greatly appreciated!
Thanks in advance!
The Laplace transform of the Delta distrubution is

$\displaystyle L\left( \delta_a(t) \right)=e^{-as}$

So taking the Transform of the equation gives

$\displaystyle s^2Y-s+1-Y=e^{-3s} \iff Y = \frac{s-1}{s^2-1}+\frac{e^{-3s}}{(s^2-1)}$

Simplifying gives and doing partial fractions gives

$\displaystyle Y=\frac{1}{s+1}+\frac{\frac{1}{2}e^{-3s}}{s-1}-\frac{\frac{1}{2}e^{-3s}}{s+1}$

Now we just take the inverse transfrom to get

$\displaystyle y(t)=e^{-t}+\frac{1}{2}e^{-(t-3)}U_3(t)-\frac{1}{2}e^{t-3}U_3(t)$

Where $\displaystyle U_3(t)$ is the unit step function shifted 3 units to the right.