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

The solution for 4y^{''}+4y^'+8y=0 is y=e^{-t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))

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

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

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

x \to 0 lim (cos (9 x))

n = 1 \sum \infty n 2^{- n}

\int x y e^{y^{2} x} d y

\frac{3}{e ^{x} + e ^{- x}}

\frac{d}{d x} (k e^{x})

What is the solution for 4y^{''}+4y^'+8y=0 ?
The solution for 4y^{''}+4y^'+8y=0 is y=e^{-t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))