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

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

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

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

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

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