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

The solution for 2y^{''}+y^'-y=0,y(0)=0,y^'(0)=2 is y= 4/3 e^{t/2}-4/3 e^{-t}

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

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

y = \frac{4}{3} e^{\frac{t}{2}} - \frac{4}{3} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{4}{3} e^{\frac{t}{2}} - \frac{4}{3} e^{- t}

y = \frac{4}{3} e^{\frac{t}{2}} - \frac{4}{3} e^{- t}

(\frac{d ^{2} y}{d x ^{2}}) - 4 y = 0

y^{^{'}} + 3 y^{^{''}} - y^{^{'}} - 3 y = 0

x^{^{''}} + 4 x = 0, x (0) = 1, x^{^{'}} (0) = - 2

y^{^{''}} + 12 y^{^{'}} + 37 y = 4 y

y^{^{''}} - y^{^{'}} - 12 y = 0, y (1) = 0 y^{^{'}} (1)

What is the solution for 2y^{''}+y^'-y=0,y(0)=0,y^'(0)=2 ?
The solution for 2y^{''}+y^'-y=0,y(0)=0,y^'(0)=2 is y= 4/3 e^{t/2}-4/3 e^{-t}