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

The solution for y^{''}-3y^'+2y=e^{-x} is y=c_{1}e^{2x}+c_{2}e^x+1/6 e^{-x}

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{6} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{6} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{6} e^{- x}

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x ^{5}}

y^{^{''}} - 4 y = 3 x e^{2 x}

y^{^{''}} - 25 y = 3 x^{2} - 2 + e^{x}

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

What is the solution for y^{''}-3y^'+2y=e^{-x} ?
The solution for y^{''}-3y^'+2y=e^{-x} is y=c_{1}e^{2x}+c_{2}e^x+1/6 e^{-x}