What is the solution for 12y^{''}+5y^'-2y=0 ?

The solution for 12y^{''}+5y^'-2y=0 is y=c_{1}e^{t/4}+c_{2}e^{-(2t)/3}

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12y^{''}+5y^'-2y=0

12 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

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

Solution steps

12 y^{''} (t) + 5 y^{'} (t) - 2 y = 0

Solve linear ODE:

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

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

3 y^{^{''}} + 30 y^{^{'}} + 14 y = 0

y^{^{''}} - 9 y = 0, y (0) = 5, y^{^{'}} (0) = 9

2 y^{^{''}} + 9 y^{^{'}} - y = 0

y^{^{''}} + \frac{1}{4} y^{^{'}} + 2 y = 0

y^{^{''}} = 3 y^{^{'}} + 5 y, y (0) = 8

What is the solution for 12y^{''}+5y^'-2y=0 ?
The solution for 12y^{''}+5y^'-2y=0 is y=c_{1}e^{t/4}+c_{2}e^{-(2t)/3}