I want to solve the homogeneous first-order ODE $$y'=\sin(y/x)+(y/x)$$ Using substitution $v=y/x$ we separate variables to get $$\csc(v)\;dv=1/x\;dx$$ and hence $$|\csc(v)+\cot(v)|^{-1}=A|x|,\;\;\;A>0.$$ But this is nasty and I don't see how to give a closed-form general solution. Maybe I made a mistake in my work above.
Ideas?
Thanks!
On
The trick is to use some kind of trigonometric identity, for instance $$ \frac{dx}{x}=\frac{dv}{\sin v}=\frac{\sin(v)\,dv}{1-\cos^2v}. $$ Now perform partial fraction decomposition and integrate $$ 2\frac{dx}{x}=\frac{\sin v\,dv}{1-\cos v}+\frac{\sin v\,dv}{1+\cos v}, \\ c+2\ln |x|=\ln|1-\cos v|-\ln|1+\cos v|, $$ to get an implicit form of the solution. There are now multiple ways to solve, one could extract $\cos v$ and then apply the inverse cosine, $$ Cx^2=\frac{1-\cos v}{1+\cos v},~~C\ge0, \\ \cos v=\frac{1-Cx^2}{1+Cx^2}, \\ v=\cos^{-1}\frac{1-Cx^2}{1+Cx^2}. $$ Or use the double angle identities to get $$ Cx^2=\tan^2\frac v2, \\ v=2\tan^{-1}(C_1x). $$
You can rewrite $$\csc v + \cot v = \cot\left(\frac{v}{2}\right)$$ so after separation of variables
You get (rewriting the constant): $$-\log\left(\cot\left(\frac{v}{2}\right)\right)=\log x + c \iff \log\left(\cot\left(\frac{v}{2}\right)\right)=\log x^{-1} - c\iff \cot\left(\frac{v}{2}\right)= \frac{a}{x}$$ This allows to explicitly solve for $v$ and via $v=y/x$ for y as well: $$\frac{v}{2}=\cot^{-1}\left(\frac{a}{x}\right)\iff y=2x\cot^{-1}\left(\frac{a}{x}\right)$$