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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 2 y = - 2 x^{3} - 3 x^{2} + 8 x + 1

y^{^{''}} - 9 y = \frac{9 x}{e ^{3 x}}

y^{^{''}} + y = 2 x sin (x)

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

y^{^{''}} - y^{^{'}} - 2 y = sin (2 x)

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