What is the solution for y^{''}-2y^'-3y=2xe^{3x} ?

The solution for y^{''}-2y^'-3y=2xe^{3x} is y=c_{1}e^{3x}+c_{2}e^{-x}+(e^{3x}x^2)/4-(e^{3x}x)/8

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y^{''}-2y^'-3y=2xe^{3x}

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

y = c_{1} e^{3 x} + c_{2} e^{- x} + \frac{e ^{3 x} x ^{2}}{4} - \frac{e ^{3 x} x}{8}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 2 x e^{3 x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} + \frac{e ^{3 x} x ^{2}}{4} - \frac{e ^{3 x} x}{8}

y = c_{1} e^{3 x} + c_{2} e^{- x} + \frac{e ^{3 x} x ^{2}}{4} - \frac{e ^{3 x} x}{8}

What is the solution for y^{''}-2y^'-3y=2xe^{3x} ?
The solution for y^{''}-2y^'-3y=2xe^{3x} is y=c_{1}e^{3x}+c_{2}e^{-x}+(e^{3x}x^2)/4-(e^{3x}x)/8