What is the solution for 2y^{''}+4y^'+8y=8 ?

The solution for 2y^{''}+4y^'+8y=8 is y=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))+1

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2y^{''}+4y^'+8y=8

2 y^{^{''}} + 4 y^{^{'}} + 8 y = 8

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

Solution steps

2 y^{''} (t) + 4 y^{'} (t) + 8 y = 8

Solve linear ODE:

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

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

What is the solution for 2y^{''}+4y^'+8y=8 ?
The solution for 2y^{''}+4y^'+8y=8 is y=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))+1