What is the solution for y^{''}+2y^'+2y=3e^{-x} ?

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

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

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

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + 3 e^{- x}

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + 3 e^{- x}

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + 3 e^{- x}

y^{^{''}} - 6 y^{^{'}} + 8 y = 3 sin (2 x)

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

y^{^{''}} + 2 y = - 4 x + 10

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (2 t)

y^{^{''}} - 6 y^{^{'}} + 5 = 0

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