Question

find the general solution of y'' + y' - 2y = 3(e^-2x) + x^2 - 3x + (e^x)(sinx)

Answer 1

\[y \prime \prime + y \prime - 2y = 3e^{-2x} + x^2 - 3x + e ^{x} sinx\]

Answer 2

hi thank you so so much for the amazing work out and answers. i'm trying to work it out again and there's one thing I don't understand which is why is there an extra x in \[y_{p1} = p _{1}xe ^{-2x}\] i thought it would just be \[y_{p1} = p_{1}e ^{-2x}\]

Answer 3

nevermind i understand why now. thanks once again for the amazing and thorough answer!