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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 49 y = sec (7 x)

\frac{d ^{2} y}{d x ^{2}} + 4 y = cos (2 x) + x e^{3 x}

y^{^{''}} - y^{^{'}} - 2 y = 10 sin (2 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = 4 e^{2 x}

y^{^{''}} - 49 y = cosh (7 x)

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