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

The solution for y^{''}-3y^'-4y=e^{-x} is y=c_{1}e^{4x}+c_{2}e^{-x}-(e^{-x}x)/5

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

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

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{e ^{- x} x}{5}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{e ^{- x} x}{5}

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{e ^{- x} x}{5}

3 y^{^{''}} + 27 y = 2 cos (3 t)

y^{^{''}} + 16 y = 32 t^{2} - 48 t + 100, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 9 y = cosh (3 x)

y^{^{''''}} - 16 y = - 12 t^{3}

What is the solution for y^{''}-3y^'-4y=e^{-x} ?
The solution for y^{''}-3y^'-4y=e^{-x} is y=c_{1}e^{4x}+c_{2}e^{-x}-(e^{-x}x)/5