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

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

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

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

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

16 y^{^{''}} + 40 y^{^{'}} + 169 y = 0, y (0) = 8, y^{^{'}} (0) = 18

(D^{4} + 6 D^{2} + 5) y = 0

x^{^{''}} + 2 x^{^{'}} + 7 x = 0

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

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