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}}

y^{^{''}} - 8 y^{^{'}} + 17 y = 0

x^{^{''}} + 4 x = 0

16 \frac{d ^{4} y}{d x ^{4}} + 24 \frac{d ^{2} y}{d x ^{2}} + 9 y = 0

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 1

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}