You can always double-check solutions to DE's yourself by plugging back the solution into the original DE and see if you get equality. It is also straight-forward to check the initial conditions.x'' + 2x' + x = t with x(0) = x'(0) = 1.
My answer is x = t^2(1-t) + t - 1 + e^-t(t+1) using variation of parameters.
x'' + 2mx' + (w^2 + m^2)x = 0 with x(0) = 0, x'(0) = A.
My answer x = A/w*e^mt*(sinwt)
Much appreciated (although I am providing good practice for other people)
I'm not sure what there is for helpers to do here.