What is the solution for 3y^{'''}-5y^{''}-16y^'+12y=0 ?

The solution for 3y^{'''}-5y^{''}-16y^'+12y=0 is y=c_{1}e^{-2t}+c_{2}e^{(2t)/3}+c_{3}e^{3t}

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3y^{'''}-5y^{''}-16y^'+12y=0

3 y^{^{'''}} - 5 y^{^{''}} - 16 y^{^{'}} + 12 y = 0

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

Solution steps

3 y^{'''} (t) - 5 y^{''} (t) - 16 y^{'} (t) + 12 y = 0

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + 4.25 y = 0, y (0) = 5

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 40 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

y^{^{''}} + 8 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = 1

y^{^{''''}} + 12 y^{^{'''}} + 36 y^{^{''}} = 0

What is the solution for 3y^{'''}-5y^{''}-16y^'+12y=0 ?
The solution for 3y^{'''}-5y^{''}-16y^'+12y=0 is y=c_{1}e^{-2t}+c_{2}e^{(2t)/3}+c_{3}e^{3t}