What is the solution for y^{''}-2y^'-8y=18te^{4t} ?

The solution for y^{''}-2y^'-8y=18te^{4t} is y=c_{1}e^{4t}+c_{2}e^{-2t}+(3e^{4t}t^2)/2-(e^{4t}t)/2

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y^{''}-2y^'-8y=18te^{4t}

y^{^{''}} - 2 y^{^{'}} - 8 y = 18 t e^{4 t}

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} + \frac{3 e ^{4 t} t ^{2}}{2} - \frac{e ^{4 t} t}{2}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = 18 t e^{4 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} + \frac{3 e ^{4 t} t ^{2}}{2} - \frac{e ^{4 t} t}{2}

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} + \frac{3 e ^{4 t} t ^{2}}{2} - \frac{e ^{4 t} t}{2}

What is the solution for y^{''}-2y^'-8y=18te^{4t} ?
The solution for y^{''}-2y^'-8y=18te^{4t} is y=c_{1}e^{4t}+c_{2}e^{-2t}+(3e^{4t}t^2)/2-(e^{4t}t)/2