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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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