What is the solution for y^{''}+20y^'+36y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+20y^'+36y=0,y(0)=1,y^'(0)=0 is y=(1+1/8)e^{-2t}-1/8 e^{-18t}

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y^{''}+20y^'+36y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 + \frac{1}{8}) e^{- 2 t} - \frac{1}{8} e^{- 18 t}

Solution steps

y^{''} (t) + 20 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = (1 + \frac{1}{8}) e^{- 2 t} - \frac{1}{8} e^{- 18 t}

y = (1 + \frac{1}{8}) e^{- 2 t} - \frac{1}{8} e^{- 18 t}

64 y^{^{''}} - 48 y^{^{'}} + 9 y = 0

x^{^{''}} - 2 x^{^{'}} + 2 x = 0

u^{^{''}} + 0.3 u^{^{'}} + 2.7 H (t) = 0

y^{^{''}} + 5 y^{^{'}} + y = 0, y (0) = 0

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 15

What is the solution for y^{''}+20y^'+36y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+20y^'+36y=0,y(0)=1,y^'(0)=0 is y=(1+1/8)e^{-2t}-1/8 e^{-18t}