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

The solution for y^{''}+y^'-2y=te^{3t} is y=c_{1}e^t+c_{2}e^{-2t}+(e^{3t}t)/(10)-7/100 e^{3t}

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

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

y = c_{1} e^{t} + c_{2} e^{- 2 t} + \frac{e ^{3 t} t}{10} - \frac{7}{100} e^{3 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 2 t} + \frac{e ^{3 t} t}{10} - \frac{7}{100} e^{3 t}

y = c_{1} e^{t} + c_{2} e^{- 2 t} + \frac{e ^{3 t} t}{10} - \frac{7}{100} e^{3 t}

y^{^{''}} + 24 y^{^{'}} + 144 y = t^{2} e^{- 12 t} + e^{- 12 t}

y^{^{''}} - 81 y = 3

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 4 y = 4 sin (2 t), y (π) = 0, y^{^{'}} (π) = - 1

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 40, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+y^'-2y=te^{3t} ?
The solution for y^{''}+y^'-2y=te^{3t} is y=c_{1}e^t+c_{2}e^{-2t}+(e^{3t}t)/(10)-7/100 e^{3t}