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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{x ^{2}}{x ^{2} + 25} d x

y^{^{'}} = 7 y^{2}

\frac{d}{d x} (\frac{x}{3 2 x})

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 6} e^{4 t}

\frac{\partial}{\partial y} (8 x^{5 y})

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