What is the solution for y^{''}+4y=2e^{-x} ?

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

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

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

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

Solution steps

y^{''} (x) + 4 y = 2 e^{- x}

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 10 y = 50 x, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 3 y^{^{'}} = \frac{10 sin ( 7 t )}{2}

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

y^{^{''}} - 4 y^{^{'}} + 3 y = \frac{e ^{t}}{1 + e ^{t}}

\frac{d ^{2} y}{d t ^{2}} + 9 y = 12 cos (3 t)

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