What is the solution for 6y^{'''}+4y^{''}=0 ?

The solution for 6y^{'''}+4y^{''}=0 is y=c_{1}e^{-(2t)/3}+c_{2}+c_{3}t

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6y^{'''}+4y^{''}=0

6 y^{^{'''}} + 4 y^{^{''}} = 0

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

Solution steps

6 y^{'''} (t) + 4 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

\int \frac{x}{e ^{x^{2}}} d x

\int \frac{1 + x ^{2}}{1 + x ^{4}} d x

\frac{\partial}{\partial y} (- x^{2} y)

f (x) = \frac{1}{x - 1}, at x = - 4

y^{^{'}} = (t y^{2} - 1) y

What is the solution for 6y^{'''}+4y^{''}=0 ?
The solution for 6y^{'''}+4y^{''}=0 is y=c_{1}e^{-(2t)/3}+c_{2}+c_{3}t