What is the solution for 8y^{''}+4y^'+y=0 ?

The solution for 8y^{''}+4y^'+y=0 is y=e^{-t/4}(c_{1}cos(t/4)+c_{2}sin(t/4))

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

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

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{t}{4}) + c_{2} sin (\frac{t}{4}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{t}{4}) + c_{2} sin (\frac{t}{4}))

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{t}{4}) + c_{2} sin (\frac{t}{4}))

What is the solution for 8y^{''}+4y^'+y=0 ?
The solution for 8y^{''}+4y^'+y=0 is y=e^{-t/4}(c_{1}cos(t/4)+c_{2}sin(t/4))