# Solving a differential equation with the Dirac-Delta function without Laplace transformations

So I'm trying to solve the following differential equation: $y''+3y'+2y=\delta(t-1)$, $y(0)=0$, $y'(0)=0$. (where $\delta$ is the Dirac's delta function) Everything I've read in my textbook/online has solved these types of equations by taking the Laplace transformation, but our class hasn't covered Laplace transformations yet...anyone have any idea what I should do?

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What did you go over in class - have you even seen these sorts of differential equations before? Were you assigned some reading material preceding the homework? Are these textbook questions that lie at the end of a chapter? It's perplexing that you'd be expected to solve an equation but not expected to learn about how to do it first... – anon Oct 7 '11 at 3:36
Well, in class we did this example: y''-y=delta(t-3). The homogeneous solutions were y=e^-t and y=e^t. He then deduced that the particular solution would be: [(e^t)/2]*integral from 0 to t ( (e^-s)*delta(s-3)ds ) - [e^(-t)/2]*integral from 0 to t( (e^s)*delta(s-3))ds From there, he concluded that y(t) would be e^(t-3)/2 - e^(3-t)/2 for all t>3. I guess what I'm confused on is how much of the partucluar solution came from the homogenous, and how much is just always going to be part of the particular solution – ben Oct 7 '11 at 3:53
And does the way he 'then deduced' the particular solution have anything to do with formulas you already have? How did he justify it to the class if not with either a derivation or a formula you're expected to know? – anon Oct 7 '11 at 4:02
Dear ben, you can use the formula given in this answer. – Pierre-Yves Gaillard Oct 7 '11 at 15:36

The physicist's answer, not worrying about convergence, uniqueness, etc. It sounds like you have no problem away from $t=1$. Up to $1^-,\ \ y(t)=0$. Above $1$, you can solve it with $y$ being a sum of exponentials. Crossing $1$, you should integrate: $\int_{1^-}^{1^+}y''+3y'=1=y'+3y|_{1^-}^{1^+}$ As $y$ can't change instantaneously, $y'$ has to go from $0$ to $1$. Then solve it starting at $t=1$ with $y=0, y'=1$