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

The solution for y^{''}-3y^'+2y=te^{-3t} is y=c_{1}e^{2t}+c_{2}e^t+(e^{-3t}t)/(20)+9/400 e^{-3t}

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

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

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{e ^{- 3 t} t}{20} + \frac{9}{400} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{e ^{- 3 t} t}{20} + \frac{9}{400} e^{- 3 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{e ^{- 3 t} t}{20} + \frac{9}{400} e^{- 3 t}

y^{^{''}} + 4 y = t^{2} + 4 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} - 4 y = 4 sin (2 x)

y^{^{''}} + y^{^{'}} = sin (2 x)

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (x) + 3 cos (2 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 cos (t)

What is the solution for y^{''}-3y^'+2y=te^{-3t} ?
The solution for y^{''}-3y^'+2y=te^{-3t} is y=c_{1}e^{2t}+c_{2}e^t+(e^{-3t}t)/(20)+9/400 e^{-3t}