What is the solution for y^{''}-y=xe^{3x} ?

The solution for y^{''}-y=xe^{3x} is y=c_{1}e^x+c_{2}e^{-x}+(e^{3x}x)/8-3/32 e^{3x}

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y^{''}-y=xe^{3x}

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

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{3 x} x}{8} - \frac{3}{32} e^{3 x}

Solution steps

y^{''} (x) - y = x e^{3 x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{3 x} x}{8} - \frac{3}{32} e^{3 x}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{e ^{3 x} x}{8} - \frac{3}{32} e^{3 x}

y^{^{'''}} + 2 y^{^{''}} + 4 y^{^{'}} + 8 y = t^{2}

y^{^{''}} + 7 y^{^{'}} + 6 y = 30

y^{^{''}} - 4 y^{^{'}} + 13 y = sin (3 x)

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

y^{^{''}} + y^{^{'}} - 12 y = - sin (2 x) + 3 cos (2 x)

What is the solution for y^{''}-y=xe^{3x} ?
The solution for y^{''}-y=xe^{3x} is y=c_{1}e^x+c_{2}e^{-x}+(e^{3x}x)/8-3/32 e^{3x}