What is the solution for y^{''}+8y^'+13y=0 ?

The solution for y^{''}+8y^'+13y=0 is y=c_{1}e^{(-4+sqrt(3))t}+c_{2}e^{(-4-sqrt(3))t}

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y^{''}+8y^'+13y=0

y^{^{''}} + 8 y^{^{'}} + 13 y = 0

y = c_{1} e^{(- 4 + 3) t} + c_{2} e^{(- 4 - 3) t}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = c_{1} e^{(- 4 + 3) t} + c_{2} e^{(- 4 - 3) t}

y = c_{1} e^{(- 4 + 3) t} + c_{2} e^{(- 4 - 3) t}

x \frac{d y}{d x} - y = x^{2} y^{2}

\frac{1}{( x ^{2} + 25 ) ^{- 3}}

y^{^{'}} + 20 y = 24

f (x) = (x^{5} - 2 x)^{2}

\int x^{2} - 9 d x

What is the solution for y^{''}+8y^'+13y=0 ?
The solution for y^{''}+8y^'+13y=0 is y=c_{1}e^{(-4+sqrt(3))t}+c_{2}e^{(-4-sqrt(3))t}