What is the solution for 4y^{''}-2y^'-2y=0,y(0)=3,y^'(0)=1 ?

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

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4y^{''}-2y^'-2y=0,y(0)=3,y^'(0)=1

4 y^{^{''}} - 2 y^{^{'}} - 2 y = 0, y (0) = 3, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{'''}} - 10 x^{^{''}} + 37 x^{^{'}} - 52 x = 0

x^{^{''}} + 6 x^{^{'}} + 8 x = 0, x (0) = 7

y^{^{''}} + 8 y^{^{'}} + 32 y = 0, y^{^{'}} 0 = - 4, y^{0} = 1

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = - 1, y^{^{'}} (0) = 8

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

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