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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} - 6 y^{^{'''}} + 9 y^{^{''}} = 0

y^{^{''}} - 9 y^{^{'}} - 52 y = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 1, y^{^{'}} (0) = 7, y (0) = y

y^{^{''''}} - 7 y^{^{''}} - 18 y = 0

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

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