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

The solution for y^{''}-4y=te^{4t} is y=c_{1}e^{2t}+c_{2}e^{-2t}+(e^{4t}t)/(12)-1/18 e^{4t}

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

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

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{4 t} t}{12} - \frac{1}{18} e^{4 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{4 t} t}{12} - \frac{1}{18} e^{4 t}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{4 t} t}{12} - \frac{1}{18} e^{4 t}

y^{^{''}} + y^{^{'}} + \frac{256}{3} x = 50 cos (3 x)

y^{^{''}} - y^{^{'}} + y = 25 sin (x)

y^{^{''}} - 3 y^{^{'}} + y = x^{2} + x

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

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{- 3 t}

What is the solution for y^{''}-4y=te^{4t} ?
The solution for y^{''}-4y=te^{4t} is y=c_{1}e^{2t}+c_{2}e^{-2t}+(e^{4t}t)/(12)-1/18 e^{4t}