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

The solution for y^{''}-y^'-20y=0,y(0)=1,y^'(0)=0 is y=(1-5/9)e^{5t}+5/9 e^{-4t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 12 y^{^{'}} + 35 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 11, y^{^{'}} (0) = 28

7 y^{^{''}} + 13 y^{^{'}} - 2 y = 0

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = e^{2 x}

What is the solution for y^{''}-y^'-20y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-y^'-20y=0,y(0)=1,y^'(0)=0 is y=(1-5/9)e^{5t}+5/9 e^{-4t}