What is the solution for 2x^5+3yy^'=0 ?

The solution for 2x^5+3yy^'=0 is y=(sqrt(-2x^6+c_{1)})/3 ,y=-(sqrt(-2x^6+c_{1)})/3

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2x^5+3yy^'=0

2 x^{5} + 3 y y^{^{'}} = 0

y = \frac{- 2 x ^{6} + c _{1}}{3}, y = - \frac{- 2 x ^{6} + c _{1}}{3}

Solution steps

2 x^{5} + 3 y y^{'} (x) = 0

Solve separable ODE:

y = \frac{- 2 x ^{6} + c _{1}}{3}, y = - \frac{- 2 x ^{6} + c _{1}}{3}

y = \frac{- 2 x ^{6} + c _{1}}{3}, y = - \frac{- 2 x ^{6} + c _{1}}{3}

\int_{- 1}^{3} (- x^{2} + 2 x + 3) d x

\int \frac{1}{e ^{x}} d x

\int_{0}^{2} (8 t i - t^{3} j + 4 t^{7} k) d t

x \to 3 - lim (6 - x)

d y = \frac{d x}{y}

What is the solution for 2x^5+3yy^'=0 ?
The solution for 2x^5+3yy^'=0 is y=(sqrt(-2x^6+c_{1)})/3 ,y=-(sqrt(-2x^6+c_{1)})/3