What is the solution for y^{''}-y^'-12y=0 ?

The solution for y^{''}-y^'-12y=0 is y=c_{1}e^{4t}+c_{2}e^{-3t}

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y^{''}-y^'-12y=0

y^{^{''}} - y^{^{'}} - 12 y = 0

y = c_{1} e^{4 t} + c_{2} e^{- 3 t}

Solution steps

y^{''} (t) - y^{'} (t) - 12 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 3 t}

y = c_{1} e^{4 t} + c_{2} e^{- 3 t}

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What is the solution for y^{''}-y^'-12y=0 ?
The solution for y^{''}-y^'-12y=0 is y=c_{1}e^{4t}+c_{2}e^{-3t}