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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 3 y = 32 (3 x + 1) e^{- 3 x}

y^{^{''}} + y^{^{'}} = cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 1 = e^{x}

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

y^{^{''}} + 9 y = 5 csc^{2} (3 x)

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