What is the solution for y^{''}+2y^'-8y=0,y(0)=0,y^'(0)=6 ?

The solution for y^{''}+2y^'-8y=0,y(0)=0,y^'(0)=6 is y=e^{2t}-e^{-4t}

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y^{''}+2y^'-8y=0,y(0)=0,y^'(0)=6

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y = e^{2 t} - e^{- 4 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = e^{2 t} - e^{- 4 t}

y = e^{2 t} - e^{- 4 t}

y^{^{''}} - 8 y + 17 y = 0

y^{^{''}} + 14 y^{^{'}} + 48 y = 0, y (0) = 9, y^{^{'}} (0) = - 66

y^{^{''''}} = 625 y

2 y^{^{''}} + 8 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 4 y + 5 y = 0

What is the solution for y^{''}+2y^'-8y=0,y(0)=0,y^'(0)=6 ?
The solution for y^{''}+2y^'-8y=0,y(0)=0,y^'(0)=6 is y=e^{2t}-e^{-4t}