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=(3-c_{2})e^{2t}+c_{2}e^{-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 - c_{2}) e^{2 t} + c_{2} e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = (3 - c_{2}) e^{2 t} + c_{2} e^{- 4 t}

y = (3 - c_{2}) e^{2 t} + c_{2} e^{- 4 t}

y^{^{''}} - 18 y^{^{'}} + 82 y = 0, y (0) = 9, y^{^{'}} (0) = 4

(D^{5} - 6 D^{4} + 9 D^{3}) y = 0

y^{^{'''}} + y^{^{''}} + 3 y^{^{'}} - 5 y = 0

y^{^{''}} + 24 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 22

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

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=(3-c_{2})e^{2t}+c_{2}e^{-4t}