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

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

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

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

y = c_{1} e^{2 x} + c_{2} + \frac{e ^{3 x} x}{3} - \frac{4}{9} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} + \frac{e ^{3 x} x}{3} - \frac{4}{9} e^{3 x}

y = c_{1} e^{2 x} + c_{2} + \frac{e ^{3 x} x}{3} - \frac{4}{9} e^{3 x}

y^{^{'''}} - 3 y^{^{''}} + 4 y = e^{2 x}

y^{^{''}} + 9 y = 3 cos (3 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 e^{- 2 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{x}

y^{^{''}} = - 3 y^{^{'}} + 2 y + 2 x + 3, y (0) = 2, y (1) = 1

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