What is the solution for 3y^{''}+2y^'+y/4 =0 ?

The solution for 3y^{''}+2y^'+y/4 =0 is y=c_{1}e^{-t/6}+c_{2}e^{-t/2}

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3y^{''}+2y^'+y/4 =0

3 y^{^{''}} + 2 y^{^{'}} + \frac{y}{4} = 0

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

Solution steps

3 y^{''} (t) + 2 y^{'} (t) + \frac{y}{4} = 0

Solve linear ODE:

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

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

What is the solution for 3y^{''}+2y^'+y/4 =0 ?
The solution for 3y^{''}+2y^'+y/4 =0 is y=c_{1}e^{-t/6}+c_{2}e^{-t/2}