Dirac Delta Function Problem

Question

In my Differential Equations class, we had the following equation on a test today:

$y''+6y'=2\delta(t)$, $y(0)=0$, $y'(0)=1$.

I got the following using Laplace Transforms (the only way I know how to do the problem):

$s^2F(S)-sy(0)-y'(0)+6sF(s)+6y(0)=2$

$s^2F(s)+6sF(s)=3$

$F(s)=\frac{3}{s^2+6s}$

$F(s)=\frac{1}{2s}-\frac{1}{2(s+6)}$

$f(t)=\frac{1}{2}-\frac{1}{2}e^{-6t}$.

Unfortunately, this does not satisfy the initial conditions, or the original equation. What is going wrong?

Answer 1

your answer is correct. just you forgot the unit step function: $F(s)=\frac{1}{2s}-\frac{1}{2(s+6)} \iff \frac{1}{2}u(t)-\frac{1}{2}e^{-6t}u(t)$

and initial condition is for $t=0^{-}$ and your answer is for $t=0^+$.