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}}

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (1) = 1, y^{^{'}} (1) = 1

y^{^{''}} + y = 0, y (0) = 19, y^{^{'}} (0) = 36

y^{^{''}} + 5 y^{^{'}} = 0, y^{1} (0) = 1

y^{^{''}} + a^{2} \cdot n^{2} \cdot π^{2} \cdot y = 0

3 y^{^{''}} - 4 y^{^{'}} = 0

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}