What is the solution for y^{''}+2y^'-3y=te^{-3t} ?

The solution for y^{''}+2y^'-3y=te^{-3t} is y=c_{1}e^t+c_{2}e^{-3t}-(e^{-3t}t^2)/8-(e^{-3t}t)/(16)

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y^{''}+2y^'-3y=te^{-3t}

y^{^{''}} + 2 y^{^{'}} - 3 y = t e^{- 3 t}

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{e ^{- 3 t} t ^{2}}{8} - \frac{e ^{- 3 t} t}{16}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 3 y = t e^{- 3 t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{e ^{- 3 t} t ^{2}}{8} - \frac{e ^{- 3 t} t}{16}

y = c_{1} e^{t} + c_{2} e^{- 3 t} - \frac{e ^{- 3 t} t ^{2}}{8} - \frac{e ^{- 3 t} t}{16}

y^{^{'''}} + 25 y^{^{'}} = 10 sin (5 x) - 4 x^{2}

y^{^{''}} + 4 y^{^{'}} + 8 y = e^{2 x} (sin (2 x) + cos (2 x))

y^{^{''}} + 2 y^{^{'}} + y = x + e^{x}

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

y^{^{''}} - 3 y + 2 y = e^{t} sin (t)

What is the solution for y^{''}+2y^'-3y=te^{-3t} ?
The solution for y^{''}+2y^'-3y=te^{-3t} is y=c_{1}e^t+c_{2}e^{-3t}-(e^{-3t}t^2)/8-(e^{-3t}t)/(16)