What is the solution for (1/8)y^{''}+4y^'+20y=0 ?

The solution for (1/8)y^{''}+4y^'+20y=0 is y=c_{1}e^{4(sqrt(6)-4)t}+c_{2}e^{-4(4+sqrt(6))t}

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(1/8)y^{''}+4y^'+20y=0

(\frac{1}{8}) y^{^{''}} + 4 y^{^{'}} + 20 y = 0

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

Solution steps

(\frac{1}{8}) y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

y = c_{1} e^{4 (6 - 4) t} + c_{2} e^{- 4 (4 + 6) t}

\int 30 (x + 9) (x + 9)^{4} d x

\frac{d}{d x} (\frac{x ^{3}}{2} - \frac{3}{x ^{2}})

y^{^{''}} - 7 y^{^{'}} + 10 y = - 2 cos (2 t)

f (x) = 4 x - 1

\frac{d y}{d x} = 8 e^{x - y}

What is the solution for (1/8)y^{''}+4y^'+20y=0 ?
The solution for (1/8)y^{''}+4y^'+20y=0 is y=c_{1}e^{4(sqrt(6)-4)t}+c_{2}e^{-4(4+sqrt(6))t}