What is the solution for y^{''}+5y^'+4y=-8te^{2t} ?

The solution for y^{''}+5y^'+4y=-8te^{2t} is y=c_{1}e^{-t}+c_{2}e^{-4t}-(4e^{2t}t)/9+2/9 e^{2t}

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y^{''}+5y^'+4y=-8te^{2t}

y^{^{''}} + 5 y^{^{'}} + 4 y = - 8 t e^{2 t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} - \frac{4 e ^{2 t} t}{9} + \frac{2}{9} e^{2 t}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = - 8 t e^{2 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} - \frac{4 e ^{2 t} t}{9} + \frac{2}{9} e^{2 t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} - \frac{4 e ^{2 t} t}{9} + \frac{2}{9} e^{2 t}

What is the solution for y^{''}+5y^'+4y=-8te^{2t} ?
The solution for y^{''}+5y^'+4y=-8te^{2t} is y=c_{1}e^{-t}+c_{2}e^{-4t}-(4e^{2t}t)/9+2/9 e^{2t}