What is the solution for y^{''}-4y^'+3y=e^{-x} ?

The solution for y^{''}-4y^'+3y=e^{-x} is y=c_{1}e^{3x}+c_{2}e^x+1/8 e^{-x}

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y^{''}-4y^'+3y=e^{-x}

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{- x}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{1}{8} e^{- x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = e^{- x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{1}{8} e^{- x}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{1}{8} e^{- x}

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What is the solution for y^{''}-4y^'+3y=e^{-x} ?
The solution for y^{''}-4y^'+3y=e^{-x} is y=c_{1}e^{3x}+c_{2}e^x+1/8 e^{-x}