What is the solution for y^{''}+13y^'+36y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+13y^'+36y=0,y(0)=1,y^'(0)=0 is y=(1+4/5)e^{-4t}-4/5 e^{-9t}

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y^{''}+13y^'+36y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 13 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 + \frac{4}{5}) e^{- 4 t} - \frac{4}{5} e^{- 9 t}

Solution steps

y^{''} (t) + 13 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = (1 + \frac{4}{5}) e^{- 4 t} - \frac{4}{5} e^{- 9 t}

y = (1 + \frac{4}{5}) e^{- 4 t} - \frac{4}{5} e^{- 9 t}

\int - 6 sin^{2} (t) d t

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

x \to \infty lim (\frac{1}{1})

n = 0 \sum \infty \frac{n + 1}{2 n - 1}

f (x) = 5 x^{4} ln (x)

What is the solution for y^{''}+13y^'+36y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+13y^'+36y=0,y(0)=1,y^'(0)=0 is y=(1+4/5)e^{-4t}-4/5 e^{-9t}