What is the solution for y^{''}+4y^'+y8=0 ?

The solution for y^{''}+4y^'+y8=0 is y=e^{-2t}(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}+4y^'+y8=0

y^{^{''}} + 4 y^{^{'}} + y 8 = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + y 8 = 0

Solve linear ODE:

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

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

What is the solution for y^{''}+4y^'+y8=0 ?
The solution for y^{''}+4y^'+y8=0 is y=e^{-2t}(c_{1}cos(2t)+c_{2}sin(2t))