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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 6 = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + 11 y = 5 e^{3 x}

y^{^{''}} + 6 y^{^{'}} + 10 y = 180 sin (2 x)

y^{^{''}} + y^{^{'}} - 72 y = - 220 t + 144 t^{2}

y^{^{''}} - 10 y^{^{'}} + 25 y = 8 e^{5 x}

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