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

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

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

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

y = (8 - c_{2}) e^{2 t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (8 - c_{2}) e^{2 t} + c_{2} e^{- t}

y = (8 - c_{2}) e^{2 t} + c_{2} e^{- t}

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y^{0} = 1

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

36 y^{^{''}} - 36 y^{^{'}} + 13 y = 0

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

y^{^{''}} + 24 y^{^{'}} - 10 y = 0

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