What is the solution for y^{''}+4y^'=xe^{4x} ?

The solution for y^{''}+4y^'=xe^{4x} is y=c_{1}+c_{2}e^{-4x}+(e^{4x}x)/(32)-3/256 e^{4x}

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y^{''}+4y^'=xe^{4x}

y^{^{''}} + 4 y^{^{'}} = x e^{4 x}

y = c_{1} + c_{2} e^{- 4 x} + \frac{e ^{4 x} x}{32} - \frac{3}{256} e^{4 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) = x e^{4 x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{e ^{4 x} x}{32} - \frac{3}{256} e^{4 x}

y = c_{1} + c_{2} e^{- 4 x} + \frac{e ^{4 x} x}{32} - \frac{3}{256} e^{4 x}

x^{^{''}} + 4 x^{^{'}} + 8 x = 4 e^{- 2 t} sin (2 t)

y^{^{''}} - 2 y^{^{'}} + 10 y = - sin (3 t)

y^{^{''''}} + y^{^{''}} = 7 x - 3 cos (x)

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

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{16 e ^{- 3 x}}{x ^{2} + 1}

What is the solution for y^{''}+4y^'=xe^{4x} ?
The solution for y^{''}+4y^'=xe^{4x} is y=c_{1}+c_{2}e^{-4x}+(e^{4x}x)/(32)-3/256 e^{4x}