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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 10 x (4 - x), y (0) = - 5, y (4) = 10

y^{^{''}} - 4 y^{^{'}} + 9 y = 2 x - 3

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 8 cos (2 x)

(D^{4} - 2 D^{2} + 1) y = cos (x)

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 x e^{2 x}

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