Solution of differential equation $\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}$ Solve the given differential equation.
$\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}$
I have been trying to solve given differential equation using elementary approaches but no manipulation is a leading to a solvable form. Could someone help me with this?
 A: $$y'(x)=\frac{1}{xy(x)\left(x^2\sin(y(x)^2)+1\right)}\Longleftrightarrow$$

Write the differential equation in terms of $x$.
Notice: $\frac{\text{d}y(x)}{\text{d}x}\cdot\frac{\text{d}x(y)}{\text{d}y}=1$

$$\frac{1}{x'(y)}=\frac{1}{yx(y)\left(x(y)^2\sin(y^2)+1\right)}\Longleftrightarrow$$
$$x'(y)=y\sin(y^2)x(y)^3+yx(y)\Longleftrightarrow$$
$$x'(y)-yx(y)=y\sin(y^2)x(y)^3\Longleftrightarrow$$
$$\frac{2y}{x(y)^2}-\frac{2x'(y)}{x(y)^3}=-2y\sin(y^2)\Longleftrightarrow$$

Let $q(y)=\frac{1}{x(y)^2}$, which gives $v'(y)=-\frac{2x'(y)}{x(y)^3}$:

$$q'(y)+2yq(y)=-2y\sin(y^2)\Longleftrightarrow$$

Let $r(y)=\exp\left[\int2y\space\text{d}y\right]=e^{y^2}$.
Multiply both sides by $r(y)$:

$$e^{y^2}q'(y)+2yq(y)e^{y^2}=-2y\sin(y^2)e^{y^2}\Longleftrightarrow$$

Substitute $2ye^{y^2}=\frac{\text{d}}{\text{d}y}\left(e^{y^2}\right)$:

$$e^{y^2}q'(y)+\frac{\text{d}}{\text{d}y}\left(e^{y^2}\right)q(y)=-2y\sin(y^2)e^{y^2}\Longleftrightarrow$$

Apply the reverse product rule:

$$\frac{\text{d}}{\text{d}y}\left(q(y)e^{y^2}\right)=-2y\sin(y^2)e^{y^2}\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}y}\left(q(y)e^{y^2}\right)\space\text{d}y=\int-2y\sin(y^2)e^{y^2}\space\text{d}y\Longleftrightarrow$$
$$q(y)e^{y^2}=-\frac{e^{y^2}\left(\sin(y^2)-\cos(y^2)\right)}{2}+\text{C}\Longleftrightarrow$$
$$q(y)=\frac{2\text{C}e^{-y^2}+\cos(y^2)-\sin(y^2)}{2}\Longleftrightarrow$$
$$\frac{1}{x(y)^2}=\frac{2\text{C}e^{-y^2}+\cos(y^2)-\sin(y^2)}{2}\Longleftrightarrow$$
$$x(y)=\pm\frac{e^{\frac{y^2}{2}}}{\sqrt{\frac{e^{y^2}\left(\cos(y^2)-\sin(y^2)\right)}{2}+\text{C}}}$$
So, we get the following solution:
$$x=\pm\frac{e^{\frac{y(x)^2}{2}}}{\sqrt{\frac{e^{y(x)^2}\left(\cos(y(x)^2)-\sin(y(x)^2)\right)}{2}+\text{C}}}$$
