What is the solution for y^{''}-4y^'-12y=te^{4t} ?

The solution for y^{''}-4y^'-12y=te^{4t} is y=c_{1}e^{6t}+c_{2}e^{-2t}-(e^{4t}t)/(12)-1/36 e^{4t}

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y^{''}-4y^'-12y=te^{4t}

y^{^{''}} - 4 y^{^{'}} - 12 y = t e^{4 t}

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{4 t} t}{12} - \frac{1}{36} e^{4 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = t e^{4 t}

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{4 t} t}{12} - \frac{1}{36} e^{4 t}

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{e ^{4 t} t}{12} - \frac{1}{36} e^{4 t}

e^{x} \frac{( 2 x + 5 ) ^{2}}{x - 4}

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What is the solution for y^{''}-4y^'-12y=te^{4t} ?
The solution for y^{''}-4y^'-12y=te^{4t} is y=c_{1}e^{6t}+c_{2}e^{-2t}-(e^{4t}t)/(12)-1/36 e^{4t}