What is the solution for 4y^{''}-2y^'-2y=0,y(0)=2,y^'(0)=1 ?

The solution for 4y^{''}-2y^'-2y=0,y(0)=2,y^'(0)=1 is y=(2-2/3)e^t+2/3 e^{-t/2}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

- x^{^{''}} = b x

9 y^{^{''}} - 30 y^{^{'}} - 24 y = 0, y (0) = 1, y^{^{'}} (0) = 0

f^{^{''}} (x) = - \frac{f ^{^{'}} ( x )}{2 ( n + 2 )}

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

u^{^{''}} + 49 H (t) = 0, H (0) = 3, u^{^{'}} (0) = - 7

What is the solution for 4y^{''}-2y^'-2y=0,y(0)=2,y^'(0)=1 ?
The solution for 4y^{''}-2y^'-2y=0,y(0)=2,y^'(0)=1 is y=(2-2/3)e^t+2/3 e^{-t/2}