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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 30 e^{2 t}

y^{^{''}} + 25 y = 14 cos (5 x) + 21 sin (5 x)

y^{^{''}} - y = t, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 4 sin (x))

y^{^{''}} - y^{^{'}} = 5 e^{2 x}

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