Personally, I have always disliked the "Laplace Transform Method" (which is what you are using although you don't say that). Problems like this can always be done more simply just using the "characterstic equation" method (the "advantage" for engineers is that with the Laplace Transform, you can just apply formulas without having to think.)
Here, the corresponding homogeneous equation is y''+ 3y'+ 2y= 0 which has characteristic equation r^2+ 3r+ 2= (r+ 2)(r+ 1)= 0 and so has general solution y(x)= C_1e^{-t}+ C_2e^{-2t}.
Now we need only find a single function that satisfies the entire equation to add to that general solution to the homogeneous equation. I would be inclined to use "undetermined coefficients" but that requires knowing the general form of the function and you might not be familiar with the derivatives of that delta function.
So instead, I will use "variation of parameters". We seek a solution of the form y(t)= u(t)e^{-t}+ v(t)e^{-2t}. There are, of course, many different choices for functions u(t) and v(t) that will work. Differentiating, y'(t)= u'(t)e^{-t}- u(t)e^{-t}+ v'(t)e^{-2t}- 2v(t)e^{-2t}. We can narrow our search among those "many different choices" and simplify the equation by requiring that u'(t)e^{-t}+ v'(t)e^{-2t}= 0. That leaves y'(t)= -u(t)e^{-t}- 2v(t)e^{-2t}.
Differentiating again, y''(t)= -u'(t)e^{-t}+ u(t)e^{-t}- 2v'(t)e^{-2t}+ 4v(t)e^{-2t}. And putting those into the equation,
y''+ 3y'+ 2y= -u'(t)e^{-t}+ u(t)e^{-t}- 2v'(t)e^{-2t}+ 4v(t)e^{-2t}+ 3(u(t)e^{-t}- 2v(t)e^{-2t})+ 2(u(t)e^{-t}+ v(t)e^{-2t})
= (-u'(t)e{-t}- 2v'(t)e^{-2t})+ u(t)(e^{-t}- 3e^{-t}+ 2e^{-t})+ v(4e^{-2t}- 6e^{-2t}+2)
=-u'(t)e^{-t}- 2v'(t)e^{-2t}= sin(t)+ \delta(t- 3\pi)
An equation involving only the two first derivatives of u and v (there are no second derivatives because of our requirement that u'(t)e^{-t}+ v'(t)e^{-2t}= 0 and no u and v themselves because e^{-t} and e^{-2t} satisfy the homogeneous equation.)
Together with the required u'(t)e^{-t}+ v'(t)e^{-2t}= 0, that gives two equations we can solve algebraically for u' and v':
u'(t)e^{-t}+ v'(t)e^{-2t}= 0
-u'(t)e^{-t}- 2v'(t)e^{-2t}= sin(t)+ \delta(t- 3\pi)
An obvious first step is to add the two equations, eliminating u':
-v'e^{-2t}= sin(t)+ \delta(t- 3\pi) so that v'= -e^{2t}sin(t)- e^{2t}\delta(t- 3\pi)
"[/itex]-e^{2t}sin(t)[/itex] can be integrated "by parts" while the integral of -e^{2t}\delta(t- 3\pi) is, of course, just -e^{6\pi}.
To find u(t), multiply the first equation by two and then add:
u'(t)e^{-t}= sin(t)+ \delta(t- 3\pi) so that
u'= e^t sin(t)+ e^t \delta(t- 3\pi)
and that can be integrated easily.
