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

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

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y^{''}+2y^'+5y=4e^{-x}cos(2x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{- x} cos (2 x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 5 y = 4 e^{- x} cos (2 x)

Solve linear ODE:

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

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

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

y^{^{''}} + 5 y^{^{'}} - 6 y = x e^{- 5 x}

y^{^{''}} + 7 y^{^{'}} + 12 y = t^{2} e^{- 3 t}

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

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

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