What is the solution for y^{''}-2y^'-48y=0 ?

The solution for y^{''}-2y^'-48y=0 is y=c_{1}e^{8t}+c_{2}e^{-6t}

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y^{''}-2y^'-48y=0

y^{^{''}} - 2 y^{^{'}} - 48 y = 0

y = c_{1} e^{8 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 48 y = 0

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- 6 t}

y = c_{1} e^{8 t} + c_{2} e^{- 6 t}

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What is the solution for y^{''}-2y^'-48y=0 ?
The solution for y^{''}-2y^'-48y=0 is y=c_{1}e^{8t}+c_{2}e^{-6t}