I was solving this differential equation, but got stuck at a certain step. Did I do something wrong, somewhere? If not, how do I go ahead and solve it? $$\left[ x\cos\frac{y}{x}+y\sin\frac{y}{x} \right]y\cdot dx =\left[ x\cos\frac{y}{x}-y\sin\frac{y}{x}\right]x\cdot dy$$$$\implies \frac{dy}{dx}= \frac{\cos\frac{y}{x}+\frac{y}{x}\sin\frac{y}{x}}{\cos\frac{y}{x}-\frac{y}{x}\sin\frac{y}{x}}\cdot\frac{y}{x}$$

Taking $y=ux$ I got $$u+x\frac{du}{dx}=\frac{u(\cos u+u\sin u)}{\cos u-u\sin u}$$$$\implies x\frac{du}{dx}=\frac{u^2\sin u + u \cos u -u \cos u + u^2\sin u}{\cos u - u\sin u}$$$$=\frac{2u^2 \sin u}{\cos u - u\sin u}$$$$\implies \frac{dx}{x}=\frac{\cos u - u\sin u}{2u^2 \sin u}\cdot du$$$$\implies \int \frac{dx}{x} = \int \frac{\cos u - u\sin u}{2u^2 \sin u}\cdot du$$$$\implies \ln |x|=\int \frac {\cot u}{2u^2}\cdot du-\frac{1}{2}\int \frac{du}{u}$$$$\implies \ln |x|+\ln |√u|=\color{red}{\int \frac {\cot u}{2u^2}\cdot du}$$

I got till there, but haven't got any further yet.

My question

  • Is there a mistake in any of my steps? If there is, please point it out, I'll be really grateful.

  • If not, could someone help with the last integral (in red)?

  • $\begingroup$ The integral you obtained at the end isn’t solvable in terms of standard mathematical functions. I haven’t taken a look at your steps, but if they’re correct, then you’re out of luck. $\endgroup$
    – Alex D
    Commented Oct 16, 2017 at 16:41
  • $\begingroup$ @IVLIVSCÆSAR Thanks, if you're not busy, could you take a look at my steps? I'd be glad to know if all my steps up to that were correct. $\endgroup$ Commented Oct 16, 2017 at 16:43
  • $\begingroup$ I don't see any mistakes.. $\endgroup$
    – Alex D
    Commented Oct 19, 2017 at 1:01
  • $\begingroup$ @IVLIVSCÆSAR Thanks a lot for sparing your time to look at my answer! I thought I'd done some mistake somewhere. $\endgroup$ Commented Oct 19, 2017 at 8:43
  • $\begingroup$ Hint: I suggest you use an integrating factor $\endgroup$
    – Alex D
    Commented Oct 20, 2017 at 19:16


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