What is the solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=26 ?

The solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=26 is y=(1-c_{2})e^{4t}+c_{2}e^{-5t}

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y^{''}+y^'-20y=0,y(0)=1,y(0)=26

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = 1, y (0) = 26

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 5 t}

Solution steps

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

Solve linear ODE:

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 5 t}

y = (1 - c_{2}) e^{4 t} + c_{2} e^{- 5 t}

2 y^{^{''}} + 8 y^{^{'}} + 6 y = 0

7 y^{^{''}} + 33 y = 0, y (0) = 4, y^{^{'}} (0) = 9

x^{^{''}} + \frac{4}{\frac{1}{16}} x = 0, x (0) = \frac{2}{3}, x^{^{'}} (0) = - \frac{4}{3}

y^{^{''}} + 4 y^{^{'}} + 24 y = 0, y (0) = 1, y^{^{'}} (0) = 0

4 y^{^{''}} + y^{^{'}} - 2 y = 0

What is the solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=26 ?
The solution for y^{''}+y^'-20y=0,y(0)=1,y(0)=26 is y=(1-c_{2})e^{4t}+c_{2}e^{-5t}