What is the solution for 4y^{''}+4y^'-8y=0,y(0)=2,y^'(0)=1 ?

The solution for 4y^{''}+4y^'-8y=0,y(0)=2,y^'(0)=1 is y=(2-1/3)e^t+1/3 e^{-2t}

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4y^{''}+4y^'-8y=0,y(0)=2,y^'(0)=1

4 y^{^{''}} + 4 y^{^{'}} - 8 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = (2 - \frac{1}{3}) e^{t} + \frac{1}{3} e^{- 2 t}

Solution steps

4 y^{''} (t) + 4 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = (2 - \frac{1}{3}) e^{t} + \frac{1}{3} e^{- 2 t}

y = (2 - \frac{1}{3}) e^{t} + \frac{1}{3} e^{- 2 t}

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 10, y^{^{'}} (0) = 10

y^{^{''}} - y^{^{'}} - 2 y = 0

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

y^{^{''}} + 6 y^{^{'}} + 18 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 k y^{^{'}} + k y = 0

What is the solution for 4y^{''}+4y^'-8y=0,y(0)=2,y^'(0)=1 ?
The solution for 4y^{''}+4y^'-8y=0,y(0)=2,y^'(0)=1 is y=(2-1/3)e^t+1/3 e^{-2t}