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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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