What is the solution for linear 12y^{''}-5y^'-3y=0 ?

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

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linear 12y^{''}-5y^'-3y=0

linear

12 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

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

Solution steps

12 y^{''} (t) - 5 y^{'} (t) - 3 y = 0

Solve linear ODE:

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

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

y^{^{''}} + \frac{49}{\frac{4}{32}} y = 0

y^{^{''}} + 9 y = 0, y^{^{'}} (3 π) = 9, y (0) = 9

(y^{^{''}}) + 9 y = 0

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 0, y (0) = 1, y (0) = 7

2 y^{^{''}} + 9 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

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