What is the solution for y^{''}+10y^'+y=0 ?

The solution for y^{''}+10y^'+y=0 is y=c_{1}e^{(-5+2sqrt(6))t}+c_{2}e^{(-5-2sqrt(6))t}

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y^{''}+10y^'+y=0

y^{^{''}} + 10 y^{^{'}} + y = 0

y = c_{1} e^{(- 5 + 26) t} + c_{2} e^{(- 5 - 26) t}

Solution steps

y^{''} (t) + 10 y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{(- 5 + 26) t} + c_{2} e^{(- 5 - 26) t}

y = c_{1} e^{(- 5 + 26) t} + c_{2} e^{(- 5 - 26) t}

\frac{d}{d x} (ln^{2} (x + 1))

\frac{d}{d x} ((x^{2} + x)^{\frac{1}{2}})

\frac{d}{d x} ((\frac{x ^{3} - 1}{2 x ^{3} + 1})^{4})

\frac{\partial}{\partial x} (2^{x y})

f (x) = \frac{2}{( 1 - x ) ^{3}}

What is the solution for y^{''}+10y^'+y=0 ?
The solution for y^{''}+10y^'+y=0 is y=c_{1}e^{(-5+2sqrt(6))t}+c_{2}e^{(-5-2sqrt(6))t}