Differential equation: $\frac{dy}{dx} = \frac1{xy({x^2 \sin(y^2)+1})}$

Question

$\dfrac{dy}{dx} = \dfrac{1}{xy({x^2 \sin(y^2)+1})}$

What would be a quick way to solve the above differential equation?

I have tried to rearrange the equation in the form of an exact differential but I am unable to do that. I get:

$2y\sin y^2 dy = \dfrac{2dx}{x^3} - \dfrac{2y}{x^2}dy$ which is not further simplifiable.

Answer 1

Possible Hint: See $$\frac{dy}{dx}=1/(\frac{dx}{dy})$$ It seems to me that this ,made your ODE to a Bernoulli's equation ODE.

Answer 2

Hint

I think that it could be easier to consider $$\dfrac{dx}{dy} = {xy({x^2 \sin(y^2)+1})}$$ Now, define $x=\frac 1 {\sqrt z}$ to make $$2 y \left(\sin \left(y^2\right)+z\right)+z'=0$$ For the homogeneous part $2yz+z'=0$, the solution is simple.