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=(1-2/3)e^{2t}+2/3 e^{-t}

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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{2}{3}) e^{2 t} + \frac{2}{3} e^{- t}

Solution steps

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

Solve linear ODE:

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

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

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=(1-2/3)e^{2t}+2/3 e^{-t}