solving ODE : $xy'=\sin(x+y)$ May I ask some hints or solution for solving $xy'=\sin(x+y)$ ??
My idea was substitution : $x+y=u$ , then it becomes $x(u'-1)=\sin u$
and still I can't approach further..
 A: Let $u=x+y$ ,
Then $y=u-x$
$y'=u'-1$
$\therefore x(u'-1)=\sin u$
$x\dfrac{du}{dx}-x=\sin u$
$x\dfrac{du}{dx}=\sin u+x$
$\dfrac{du}{dx}=\dfrac{\sin u}{x}+1$
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=223:
Let $v=\tan\dfrac{u}{2}$ ,
Then $\dfrac{dv}{dx}=\dfrac{v^2}{2}+\dfrac{v}{x}+\dfrac{1}{2}$
Let $v=-\dfrac{2}{w}\dfrac{dw}{dx}$ ,
Then $\dfrac{dv}{dx}=-\dfrac{2}{w}\dfrac{d^2w}{dx^2}+\dfrac{2}{w^2}\left(\dfrac{dw}{dx}\right)^2$
$\therefore-\dfrac{2}{w}\dfrac{d^2w}{dx^2}+\dfrac{2}{w^2}\left(\dfrac{dw}{dx}\right)^2=\dfrac{2}{w^2}\left(\dfrac{dw}{dx}\right)^2-\dfrac{2}{xw}\dfrac{dw}{dx}+\dfrac{1}{2}$
$\dfrac{2}{w}\dfrac{d^2w}{dx^2}-\dfrac{2}{xw}\dfrac{dw}{dx}+\dfrac{1}{2}=0$
$4x\dfrac{d^2w}{dx^2}-4\dfrac{dw}{dx}+xw=0$
$w=C_1xJ_1\left(\dfrac{x}{2}\right)+C_2xY_1\left(\dfrac{x}{2}\right)$ (according to http://www.wolframalpha.com/input/?i=4xw''-4w'%2Bxw%3D0)
$\therefore v=-\dfrac{2\dfrac{d}{dx}\left(C_1xJ_1\left(\dfrac{x}{2}\right)+C_2xY_1\left(\dfrac{x}{2}\right)\right)}{C_1xJ_1\left(\dfrac{x}{2}\right)+C_2xY_1\left(\dfrac{x}{2}\right)}=-\dfrac{C_1xJ_0\left(\dfrac{x}{2}\right)-C_1xJ_2\left(\dfrac{x}{2}\right)+4C_1J_1\left(\dfrac{x}{2}\right)+C_2xY_0\left(\dfrac{x}{2}\right)-C_2xY_2\left(\dfrac{x}{2}\right)+4C_2Y_1\left(\dfrac{x}{2}\right)}{2C_1xJ_1\left(\dfrac{x}{2}\right)+2C_2xY_1\left(\dfrac{x}{2}\right)}=\dfrac{xJ_2\left(\dfrac{x}{2}\right)-xJ_0\left(\dfrac{x}{2}\right)-4J_1\left(\dfrac{x}{2}\right)+CxY_2\left(\dfrac{x}{2}\right)-CxY_0\left(\dfrac{x}{2}\right)-4CY_1\left(\dfrac{x}{2}\right)}{2xJ_1\left(\dfrac{x}{2}\right)+2CxY_1\left(\dfrac{x}{2}\right)}$
$u=2\tan^{-1}\dfrac{xJ_2\left(\dfrac{x}{2}\right)-xJ_0\left(\dfrac{x}{2}\right)-4J_1\left(\dfrac{x}{2}\right)+CxY_2\left(\dfrac{x}{2}\right)-CxY_0\left(\dfrac{x}{2}\right)-4CY_1\left(\dfrac{x}{2}\right)}{2xJ_1\left(\dfrac{x}{2}\right)+2CxY_1\left(\dfrac{x}{2}\right)}$
$y=2\tan^{-1}\dfrac{xJ_2\left(\dfrac{x}{2}\right)-xJ_0\left(\dfrac{x}{2}\right)-4J_1\left(\dfrac{x}{2}\right)+CxY_2\left(\dfrac{x}{2}\right)-CxY_0\left(\dfrac{x}{2}\right)-4CY_1\left(\dfrac{x}{2}\right)}{2xJ_1\left(\dfrac{x}{2}\right)+2CxY_1\left(\dfrac{x}{2}\right)}-x$
