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

The solution for y^{''}+y^'-6y=xe^{2x} is y=c_{1}e^{2x}+c_{2}e^{-3x}+(e^{2x}x^2)/(10)-(e^{2x}x)/(25)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{e ^{2 x} x ^{2}}{10} - \frac{e ^{2 x} x}{25}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{e ^{2 x} x ^{2}}{10} - \frac{e ^{2 x} x}{25}

y = c_{1} e^{2 x} + c_{2} e^{- 3 x} + \frac{e ^{2 x} x ^{2}}{10} - \frac{e ^{2 x} x}{25}

\frac{d ^{2} y}{d x ^{2}} + y = sin (x)

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 24

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

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

y^{^{''}} + 5 y^{^{'}} + 6 y = 24 e^{t} + 36 t^{2}

What is the solution for y^{''}+y^'-6y=xe^{2x} ?
The solution for y^{''}+y^'-6y=xe^{2x} is y=c_{1}e^{2x}+c_{2}e^{-3x}+(e^{2x}x^2)/(10)-(e^{2x}x)/(25)