What is the solution for y^{''}+2y^'+2y=6 ?

The solution for y^{''}+2y^'+2y=6 is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+3

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y^{''}+2y^'+2y=6

y^{^{''}} + 2 y^{^{'}} + 2 y = 6

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 3

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 6

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 3

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 3

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = x^{2} + 4 x + 8

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 8

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 6 + x - 1 e x

y^{^{''}} - 8 y^{^{'}} + 16 y = 28 x + 4

y^{^{''}} + 4 y^{^{'}} + 13 y = cos (t)

What is the solution for y^{''}+2y^'+2y=6 ?
The solution for y^{''}+2y^'+2y=6 is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+3