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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} + y = 4, y^{^{'}} (0) = - 1, y (0) = 4

y^{^{''}} + y - 2 y = 2 + 5 cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{- t}

y^{^{''}} - 2 y - 4 x^{2} e^{2 x} = 0

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 3, y^{^{'}} (0) = 1

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