What is the solution for y^{''}+6y^'+9y=xe^{2x} ?

The solution for y^{''}+6y^'+9y=xe^{2x} is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+(e^{2x}x)/(25)-2/125 e^{2x}

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y^{''}+6y^'+9y=xe^{2x}

y^{^{''}} + 6 y^{^{'}} + 9 y = x e^{2 x}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{e ^{2 x} x}{25} - \frac{2}{125} e^{2 x}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = x e^{2 x}

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{e ^{2 x} x}{25} - \frac{2}{125} e^{2 x}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{e ^{2 x} x}{25} - \frac{2}{125} e^{2 x}

x^{^{''}} + 4 x^{^{'}} + 6 x = 1 + e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 sin (2 x) - 3 e^{2 x}

y^{^{''}} + y = cos (a t)

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = - 8 e^{x} + 6 e^{- x}

y^{^{''}} + 2 y^{^{'}} + y = 6 e^{- t}, y (0) = y

What is the solution for y^{''}+6y^'+9y=xe^{2x} ?
The solution for y^{''}+6y^'+9y=xe^{2x} is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+(e^{2x}x)/(25)-2/125 e^{2x}