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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} + 25 y = 5 sin (5 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{t} cos (t)

y^{^{'''}} + 36 y^{^{'}} = sec (6 x)

y^{^{''}} + y = - cos (2 t)

y^{^{''}} - 4 y = e^{x} cos (x)

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