What is the solution for y^{''}+6y^'+9y=xe^{4x} ?

The solution for y^{''}+6y^'+9y=xe^{4x} is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+(e^{4x}x)/(49)-2/343 e^{4x}

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y^{''}+6y^'+9y=xe^{4x}

y^{^{''}} + 6 y^{^{'}} + 9 y = x e^{4 x}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{e ^{4 x} x}{49} - \frac{2}{343} e^{4 x}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = x e^{4 x}

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{e ^{4 x} x}{49} - \frac{2}{343} e^{4 x}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{e ^{4 x} x}{49} - \frac{2}{343} e^{4 x}

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

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

y^{^{''}} + 2 y^{^{'}} + y = ln (x) \cdot e^{- x}

x^{^{''}} - 14 x^{^{'}} + 49 x = 5 t e^{7 t}

y^{^{''}} + 16 y = cos (4 x)

What is the solution for y^{''}+6y^'+9y=xe^{4x} ?
The solution for y^{''}+6y^'+9y=xe^{4x} is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+(e^{4x}x)/(49)-2/343 e^{4x}