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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + (440 \cdot 2 \cdot π)^{2} y = 1

y^{^{''}} + 25 y = 5 x, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e x - 3

y^{^{''}} + 2 y^{^{'}} + 4 y = 6 cos (3 x)

x^{^{''}} + 4 x^{^{'}} + 3 x = 1, x^{^{'}} (0) = 0

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