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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = - 13

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

16 y^{^{''}} + 522.67 y = 0

\frac{d ^{2} y}{d t ^{2}} + 64 y = 0

y^{^{'''}} + 6 y^{^{''}} - 37 y^{^{'}} - 90 y = 0

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