What is the solution for y^{''}+8y^'+6y=0 ?

The solution for y^{''}+8y^'+6y=0 is y=c_{1}e^{(-4+sqrt(10))t}+c_{2}e^{(-4-sqrt(10))t}

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y^{''}+8y^'+6y=0

y^{^{''}} + 8 y^{^{'}} + 6 y = 0

y = c_{1} e^{(- 4 + 10) t} + c_{2} e^{(- 4 - 10) t}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = c_{1} e^{(- 4 + 10) t} + c_{2} e^{(- 4 - 10) t}

y = c_{1} e^{(- 4 + 10) t} + c_{2} e^{(- 4 - 10) t}

\frac{4 s + 3}{s ( s ^{2} + 6 s + 34 )}

f^{^{'}} (x) = x + 1

\frac{d}{d x} (- sin (2 x) \cdot 2 + cos (2 x) \cdot 2)

\int x^{2} - 7 x + 8 d x

- 3 x^{2} + 6 x + 4

What is the solution for y^{''}+8y^'+6y=0 ?
The solution for y^{''}+8y^'+6y=0 is y=c_{1}e^{(-4+sqrt(10))t}+c_{2}e^{(-4-sqrt(10))t}