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

The solution for y^{''}-y^'-20y=0,y(0)=1,y^'(0)=-1 is y=(1-2/3)e^{5t}+2/3 e^{-4t}

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

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

y = (1 - \frac{2}{3}) e^{5 t} + \frac{2}{3} e^{- 4 t}

Solution steps

y^{''} (t) - y^{'} (t) - 20 y = 0

Solve linear ODE:

y = (1 - \frac{2}{3}) e^{5 t} + \frac{2}{3} e^{- 4 t}

y = (1 - \frac{2}{3}) e^{5 t} + \frac{2}{3} e^{- 4 t}

6 f^{^{''}} (t) = 3 f^{^{''}} (t) - 5 f (t)

y^{^{''}} - 6 y^{^{'}} + 16 y = 0

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = 5

3 y^{^{''}} + 2 y^{^{'}} + \frac{1}{4} y = 0

(D - 1) (D + 1) (D - 2) x = 0

What is the solution for y^{''}-y^'-20y=0,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}-y^'-20y=0,y(0)=1,y^'(0)=-1 is y=(1-2/3)e^{5t}+2/3 e^{-4t}