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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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