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

The solution for y^{''}+2y^'-8y=0 is y=c_{1}e^{2t}+c_{2}e^{-4t}

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y^{''}+2y^'-8y=0

y^{^{''}} + 2 y^{^{'}} - 8 y = 0

y = c_{1} e^{2 t} + c_{2} e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 4 t}

y = c_{1} e^{2 t} + c_{2} e^{- 4 t}

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What is the solution for y^{''}+2y^'-8y=0 ?
The solution for y^{''}+2y^'-8y=0 is y=c_{1}e^{2t}+c_{2}e^{-4t}