What is the solution for y^{''}-4y-4y=0 ?

The solution for y^{''}-4y-4y=0 is y=c_{1}e^{2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}

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y^{''}-4y-4y=0

y^{^{''}} - 4 y - 4 y = 0

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

Solution steps

y^{''} (t) - 4 y - 4 y = 0

Solve linear ODE:

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

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

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

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

(2 \cdot y^{^{''}}) - (3 \cdot y^{^{'}}) - (2 \cdot y) = 0

y^{^{''}} + 20.02 y = 0, y (0) = 0, y^{^{'}} (0) = - 2

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 8

What is the solution for y^{''}-4y-4y=0 ?
The solution for y^{''}-4y-4y=0 is y=c_{1}e^{2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}