What is the solution for y^{''}+8y^'+12=0 ?

The solution for y^{''}+8y^'+12=0 is y=c_{1}+c_{2}e^{-8t}-(3t)/2

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y^{''}+8y^'+12=0

y^{^{''}} + 8 y^{^{'}} + 12 = 0

y = c_{1} + c_{2} e^{- 8 t} - \frac{3 t}{2}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 12 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 8 t} - \frac{3 t}{2}

y = c_{1} + c_{2} e^{- 8 t} - \frac{3 t}{2}

What is the solution for y^{''}+8y^'+12=0 ?
The solution for y^{''}+8y^'+12=0 is y=c_{1}+c_{2}e^{-8t}-(3t)/2