What is the solution for y^{''}+20y^'+64y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+20y^'+64y=0,y(0)=1,y^'(0)=0 is y=(1+1/3)e^{-4t}-1/3 e^{-16t}

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y^{''}+20y^'+64y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 20 y^{^{'}} + 64 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 + \frac{1}{3}) e^{- 4 t} - \frac{1}{3} e^{- 16 t}

Solution steps

y^{''} (t) + 20 y^{'} (t) + 64 y = 0

Solve linear ODE:

y = (1 + \frac{1}{3}) e^{- 4 t} - \frac{1}{3} e^{- 16 t}

y = (1 + \frac{1}{3}) e^{- 4 t} - \frac{1}{3} e^{- 16 t}

\frac{1}{8} x^{^{''}} + 36 x = 0

\frac{d ^{2} y}{d x ^{2}} = - 4 y

\frac{y ^{^{''}}}{8} + 2 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} + 9 y = 0, y^{^{'}} (\frac{π}{6}) = - 3, y (0) = 1

\frac{d ^{2} y}{d t ^{2}} - y = 0

What is the solution for y^{''}+20y^'+64y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+20y^'+64y=0,y(0)=1,y^'(0)=0 is y=(1+1/3)e^{-4t}-1/3 e^{-16t}