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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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