Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$ Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$
i am trying to solving integrating combination
since given that 
$\cot (x^2+y^2)dy+xdx+ydy=0$
then $\cot (x^2+y^2)dy+d(xy)=0$ is it correct way ? and we can apply integration from here? can any one help me this problem
 A: $$2\cot(x^2+y^2)dy+2xdx+2ydy=0$$
$$2\cot(x^2+y^2)dy+d(x^2+y^2)=0$$
$$\dfrac{d(x^2+y^2)}{\cot(x^2+y^2)}=-2dy$$
$$-\ln\cos(x^2+y^2)=-2y+C$$
A: Using polar coordinates, we see that
\begin{align}
dx =&\ \cos\theta dr -r\sin\theta d\theta,\\
dy =&\ \sin\theta dr + r\cos\theta d\theta
\end{align}
then we see that
\begin{align}
\cot(x^2+y^2)dy +xdx+ydy = (\cot(r^2)\sin\theta+r) dr+r\cot(r^2)\cos\theta d\theta =0.
\end{align}
Hence it follows
\begin{align}
\frac{d\theta}{dr} = - \frac{\cot(r^2)\sin\theta+ r}{r\cot(r^2)\cos\theta} = -\frac{\tan\theta}{r}-\tan(r^2)\sec\theta.
\end{align}
Lastly, using the fact that
\begin{align}
\frac{d (\sin\theta)}{dr} = \cos\theta\frac{d\theta}{dr} = -\frac{\sin\theta}{r}-\tan(r^2)
\end{align}
then we have obtain a linear equation
\begin{align}
u'+\frac{1}{r}u=-\tan(r^2) \ \ &\implies\ \ \frac{d}{dr}(ru) = -r\tan(r^2)\\
&\implies\  u= \frac{1}{2r}\log|\cos(r^2)|+\frac{c}{r}.
\end{align}
Finally, we have that
\begin{align}
\sin \theta =\frac{1}{2r}\log|\cos(r^2)|+\frac{c}{r} \ \implies \ y = \frac{1}{2}\log|\cos(x^2+y^2)|+C
\end{align}
A: Dividing both sides of $\cot (x^2+y^2)dy+xdx+ydy=0$ by $dy$ we get $\cot (x^2+y^2)+x\frac {dx}{dy}+y=0$ 
Then, 
$$x\frac {dx}{dy}=-y-\cot (x^2+y^2) \iff \frac {dx}{dy}=\frac{-y-\cot (x^2+y^2)}{x} \iff \frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$$
Now you can integrate this: $\frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$
A: $$\cot (x^2+y^2)dy+xdx+ydy=0$$
$$\cot (x^2+y^2)y'+x+yy'=0$$
Susbstitute $u=x^2+y^2$ and $u'=2x+2yy'$
$$\cot(u)y'+\frac {u'}2=0$$
$$\cot(u)y'=-\frac {u'}2$$
Multiply by $dx$
$$\cot(u)dy=-\frac {du}2$$
$$-2dy=\frac {du}{\cot(u)}$$
Integrate
$$-2y+K=\int\frac {du}{\cot(u)}$$
$$-2y+K=\int\frac {\sin(u)du}{\cos(u)}=-\ln|\cos(u)|$$
$$\boxed{y=\frac 12 \ln|\cos(x^2+y^2)|+K}$$
