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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 0

y^{^{''}} + 2 y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0

x^{^{''}} - 0.6 x^{^{'}} + 1.09 x = 0

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