What is the solution for y^{''}+y^'-2y=0,y(0)=1 ?

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

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y^{''}+y^'-2y=0,y(0)=1

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 1

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

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 0

Solve linear ODE:

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

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

y^{^{''}} + \frac{1}{10} y^{^{'}} = 0

y^{^{'''}} + 23 y^{^{''}} + 137 y^{^{'}} + 115 y = 0

\frac{1}{5} y^{^{''}} + 2 y^{^{'}} + 10 y = 0

81 y^{^{''''}} = y

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y^{^{'}} (0) = 3

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