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

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

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

y^{^{''}} + \frac{y ^{^{'}}}{2} - \frac{y}{2} = 0

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{3} y}{d x ^{3}} - 8 y = 0

y^{^{''''}} - 24 y^{^{'''}} + 144 y^{^{''}} = 0

(D^{4} - D^{3} - 7 D^{2} + 3 D) y = 0

3 y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 6, y^{^{'}} (0) = - 18

y^{^{''}} + 42 y^{^{'}} + 8 y = 0

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