What is the solution for 3y^{'''}+2y^{''}=0 ?

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

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3y^{'''}+2y^{''}=0

3 y^{^{'''}} + 2 y^{^{''}} = 0

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

Solution steps

3 y^{'''} (t) + 2 y^{''} (t) = 0

Solve linear ODE:

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

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

2 y^{^{''}} + 28 y^{^{'}} + 99 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 14 y^{^{'}} + 45 y = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 5, y^{^{'}} (0) = 28

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 13

16 y^{^{'''}} + 24 y^{^{''}} + 9 y^{^{'}} = 0

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