What is the solution for y^{''}-y^'-2y=te^{2t} ?

The solution for y^{''}-y^'-2y=te^{2t} is y=c_{1}e^{2t}+c_{2}e^{-t}+(e^{2t}t^2)/6-(e^{2t}t)/9

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y^{''}-y^'-2y=te^{2t}

y^{^{''}} - y^{^{'}} - 2 y = t e^{2 t}

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{e ^{2 t} t ^{2}}{6} - \frac{e ^{2 t} t}{9}

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = t e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{e ^{2 t} t ^{2}}{6} - \frac{e ^{2 t} t}{9}

y = c_{1} e^{2 t} + c_{2} e^{- t} + \frac{e ^{2 t} t ^{2}}{6} - \frac{e ^{2 t} t}{9}

- \frac{1}{9} (x^{- 9} - x^{18})

\int ln (9 x + 1) d x

\frac{d}{d x} (3 x + 8)

t (\frac{d y}{d t}) = 2 t e^{t} - y + 6 t^{2}

\int_{0}^{2 π} sin^{2} (t) d t

What is the solution for y^{''}-y^'-2y=te^{2t} ?
The solution for y^{''}-y^'-2y=te^{2t} is y=c_{1}e^{2t}+c_{2}e^{-t}+(e^{2t}t^2)/6-(e^{2t}t)/9