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

The solution for y^{''}+8y^'+4y=0 is y=c_{1}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = \frac{x ^{2}}{x ^{2} + 7}

\int - \frac{ln ( x )}{2} d x

u^{'} (t)

\frac{\partial}{\partial y} (arctan (x - y))

x \to \infty + lim (0)

What is the solution for y^{''}+8y^'+4y=0 ?
The solution for y^{''}+8y^'+4y=0 is y=c_{1}e^{2(sqrt(3)-2)t}+c_{2}e^{-2(2+sqrt(3))t}