What is the solution for y^{''}+2y^'+2y=4cos(t)+2sin(t) ?

The solution for y^{''}+2y^'+2y=4cos(t)+2sin(t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+2sin(t)

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y^{''}+2y^'+2y=4cos(t)+2sin(t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 4 cos (t) + 2 sin (t)

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 4 cos (t) + 2 sin (t)

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} + 25 y = 100 sinh (5 x)

y^{^{''}} + 2 y^{^{'}} + y = 9 e^{- t}

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

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

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

What is the solution for y^{''}+2y^'+2y=4cos(t)+2sin(t) ?
The solution for y^{''}+2y^'+2y=4cos(t)+2sin(t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+2sin(t)