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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = \frac{e ^{x}}{x ^{2}}

\int - sin^{2} (x) cos (x) d x

\frac{d}{d x} (\frac{x + 3}{x})

3 x e^{- x}

\frac{d}{d x} (5^{(6 x^{2} - 7 x) (8 x^{2} + 9)})

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