What is the solution for y^{''}+2y^'-8y=0,y(0)=3,y^'(0)=-12 ?

The solution for y^{''}+2y^'-8y=0,y(0)=3,y^'(0)=-12 is y=3e^{-4t}

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y^{''}+2y^'-8y=0,y(0)=3,y^'(0)=-12

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 3, y^{^{'}} (0) = - 12

y = 3 e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = 3 e^{- 4 t}

y = 3 e^{- 4 t}

3 y^{^{'''}} - 12 y^{^{'}} - 4 y^{^{''}} + y^{^{''''}} = 0

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

y^{^{''}} - 6 y^{^{'}} - 2 y = 0

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

y^{^{''}} + 4 y = 0, y (\frac{π}{4}) = 3, y (0) = 5

What is the solution for y^{''}+2y^'-8y=0,y(0)=3,y^'(0)=-12 ?
The solution for y^{''}+2y^'-8y=0,y(0)=3,y^'(0)=-12 is y=3e^{-4t}