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

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

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{5} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{5} e^{- x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{5} e^{- x}

\frac{\partial}{\partial x} (t^{2} e^{- x})

\int 2 x (x^{2} + 1)^{5} d x

\frac{\partial}{\partial x} (ln (5 x))

\frac{d}{d x} ((x^{3} - 4) arcsin (x))

\frac{d y}{d x} + x e^{x + y} = 0

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