(xdx -ydy) y sin y/x = ( ydx + xdx) x cos y/x

1.	(xdx -ydy) y sin y/x = ( ydx + xdx) x cos y/x
Answer» Correct QUESTION:- SOLVE the differential EQUATION $(x\ dy-y\ dx)y\sin\left(\dfrac{y}{x}\right)=(y\ dx+x\ dy)x\cos\left(\dfrac{y}{x}\right).$ Solution:- We're GIVEN to solve the differential equation, $\longrightarrow(x\ dy-y\ dx)y\sin\left(\dfrac{y}{x}\right)=(y\ dx+x\ dy)x\cos\left(\dfrac{y}{x}\right)$ Dividing by $x^2$ we get, $\longrightarrow\left(\dfrac{x\ dy-y\ dx}{x^2}\right)y\sin\left(\dfrac{y}{x}\right)=\left(\dfrac{y}{x}\ dx+dy\right)\cos\left(\dfrac{y}{x}\right)$ Put, $\longrightarrow u=\dfrac{y}{x}$ $\longrightarrow du=\dfrac{x\ dy-y\ dx}{x^2}$ Also, $\longrightarrow y=ux$ $\longrightarrow dy=u\ dx+x\ du$ Then our equation BECOMES, $\longrightarrow ux\sin u\ du=\left(2u\ dx+x\ du\right)\cos u$ $\longrightarrow ux\sin u\ du=2u\cos u\ dx+x\cos u\ du$ $\longrightarrow ux\sin u\ du-x\cos u\ du=2u\cos u\ dx$ $\longrightarrow(u\sin u-\cos u)x\ du=2u\cos u\ dx$ $\longrightarrow\dfrac{dx}{x}=\left(\dfrac{u\sin u-\cos u}{2u\cos u}\right)\ du$ $\longrightarrow\dfrac{dx}{x}=\dfrac{1}{2}\left(\tan u-\dfrac{1}{u}\right)\ du$ Now integrating, $\displaystyle\longrightarrow\int\dfrac{dx}{x}=\dfrac{1}{2}\int\left(\tan u-\dfrac{1}{u}\right)\ du$ $\displaystyle\longrightarrow\log\|x\|=\dfrac{1}{2}\left(\log\|\sec u\|-\log\|u\|\right)+\log\|C_1\|$ $\displaystyle\longrightarrow\log\|x\|=\dfrac{1}{2}\log\left\|\dfrac{\sec u}{u}\right\|+\log\|C_1\|$ Taking antilog, $\displaystyle\longrightarrow x=C_1\sqrt{\dfrac{\sec u}{u}}$ $\displaystyle\longrightarrow x^2=\dfrac{(C_1)^2\sec u}{u}$ $\displaystyle\longrightarrow \sec u=\dfrac{ux^2}{(C_1)^2}$ Taking $(C_1)^2=\dfrac{1}{C},$ $\displaystyle\longrightarrow \sec u=Cux^2$ Undoing substitution $u=\dfrac{y}{x},$ $\displaystyle\longrightarrow\underline{\underline{\sec\left(\dfrac{y}{x}\right)=Cxy}}$

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