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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} = 2 x + 1

y^{^{''}} + 3 y^{^{'}} + 2 y = 24 e^{- 2 x}

y^{^{''}} + 100 y = - 2 sin (10 t) - 4 cos (10 t)

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 5 H (t)

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