Solving Differential Equation $y'=\cos(xy)$ My question is about solving differential equation $y'=\cos(xy)$.
I tried to changing variable $u=xy$
\begin{align*}
u &= xy \\
\frac{du}{dx} &= y+ x\frac{dy}{dx}\\
\frac{dy}{dx} &= \frac{1}{x}\frac{du}{dx}-\frac{y}{x}\\
\end{align*}
then since $u=xy\to \frac{y}{x}=\frac{u}{x^2}$
\begin{align*}
\frac{1}{x}\frac{du}{dx}-\frac{u}{x^2}-\cos(u)=\;0\\
\end{align*}
finally
\begin{align*}
x\,du-(u+x^2\cos(u))\,dx=0\\
\end{align*}
since this is not an Exact Differential Equation$($Because of the existence of $\cos(u))$, what is multiplicative "integrating" factor? Is this equation solvable or not?
 A: HINT:
$$ \log ( y')= \log (\cos (xy)) $$
differentiate
$$ y''= -\sqrt{1-y^{'2}}(xy'+y)$$
Once again differentiate
$$y^{'''}\sqrt{1-y^{'2}}= xy''(1+y{'2})+y' (xy'+y+2+2 y^{'2})$$
May be no closed form solution. Numerical integration ignoring spurious solutions.
A: Hint:
Let $u=xy$ ,
Then $y=\dfrac{u}{x}$
$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$
$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\cos u$
$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{x^2\cos u+u}{x^2}$
$(x^2\cos u+u)\dfrac{dx}{du}=x$
Let $v=x^2$ ,
$\dfrac{dv}{du}=2x\dfrac{dx}{du}$
$\therefore\dfrac{(x^2\cos u+u)}{2x}\dfrac{dv}{du}=x$
$(x^2\cos u+u)\dfrac{dv}{du}=2x^2$
$(v\cos u+u)\dfrac{dv}{du}=2v$
This belongs to an Abel equation of the second kind.
Let $w=v+u\sec u$ ,
Then $v=w-u\sec u$
$\dfrac{dv}{du}=\dfrac{dw}{du}-(u\tan u+1)\sec u$
$\therefore(\cos u)w\left(\dfrac{dw}{du}-(u\tan u+1)\sec u\right)=2(w-u\sec u)$
$(\cos u)w\dfrac{dw}{du}-(u\tan u+1)w=2w-2u\sec u$
$(\cos u)w\dfrac{dw}{du}=(u\tan u+3)w-2u\sec u$
$w\dfrac{dw}{du}=(u\tan u\sec u+3\sec u)w-2u\sec^2u$
