What is the solution for ((3y^2-t^2)/(y^5))(dy)/(dt)+t/(2y^4)=0 ?

The solution for ((3y^2-t^2)/(y^5))(dy)/(dt)+t/(2y^4)=0 is

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((3y^2-t^2)/(y^5))(dy)/(dt)+t/(2y^4)=0

(\frac{3 y ^{2} - t ^{2}}{y ^{5}}) \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

y = \frac{- 3 + 2 c _{1} t ^{2} + 9}{2 c _{1}}, y = - \frac{- 3 + 2 c _{1} t ^{2} + 9}{2 c _{1}}, y = - \frac{2 c _{1} t ^{2} + 9 + 3}{2 c _{1}}, y = - - \frac{2 c _{1} t ^{2} + 9 + 3}{2 c _{1}}

Solution steps

(\frac{3 y ^{2} - t ^{2}}{y ^{5}}) \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

Solve exact ODE:

y = \frac{- 3 + 2 c _{1} t ^{2} + 9}{2 c _{1}}, y = - \frac{- 3 + 2 c _{1} t ^{2} + 9}{2 c _{1}}, y = - \frac{2 c _{1} t ^{2} + 9 + 3}{2 c _{1}}, y = - - \frac{2 c _{1} t ^{2} + 9 + 3}{2 c _{1}}

y = \frac{- 3 + 2 c _{1} t ^{2} + 9}{2 c _{1}}, y = - \frac{- 3 + 2 c _{1} t ^{2} + 9}{2 c _{1}}, y = - \frac{2 c _{1} t ^{2} + 9 + 3}{2 c _{1}}, y = - - \frac{2 c _{1} t ^{2} + 9 + 3}{2 c _{1}}

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x \to - 1 lim (1)

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What is the solution for ((3y^2-t^2)/(y^5))(dy)/(dt)+t/(2y^4)=0 ?
The solution for ((3y^2-t^2)/(y^5))(dy)/(dt)+t/(2y^4)=0 is