Differential Equations: Variation of Parameters.

y''+4y'+4y=t^ (-2) e^ (-2t)

I'd multiply by e^ (-2t) : e^ (-2t) y" - 4e^ (-2t) y' + 4e^ (-2t) y = t but e^ (-2t) y" - 4e^ (-2t) y' + 4e^ (-2t) y = (e^ (-2t) y" - 2e^ (-2t) y') + ( - 2e^ (-2t) y' + 4e^ (-2t) y) = (e^ (-2t) y ') ' - 2 (e^ (-2t) y) ' = (e^ (-2t) y ' - 2 e^ (-2t) y) ' = (e^ (-2t) y) " So the differential equation becomes: (e^ (-2t) y) " = t and so: (e^ (-2t) y) ' = (1/2) t^2 + a (a is constant) e^ (-2t) y = (1/6) t^3 + at + b (a and b constants)

so the general solution is: y = (1/6) t^3 e^ (2t) + a t e^ (2t) + be^ (2t)

the last two terms are your homogenous solution, the first term is the particular one.

Note that even if the right side was different from te^ (2t), I'd still multiply by e^ (-2t), since this is what makes the left side become (e^ (-2t) y) ", so that we just have to anti-differentiate two times.

So if we had y"-4y'+4y=t^5 instead, then the problem would become: (e^ (-2t) y) " = t^5 e^ (-2t) from which we'd find the solution by finding the primitive of the right side twice in a row, and then multiplying by e^ (2t) to get the general solution (again, the constants of integration will give the homogenous part).