Question

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

Answer 1

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 $u=\dfrac{y}{x},$

$\displaystyle\longrightarrow\underline{\underline{\sec\left(\dfrac{y}{x}\right)=Cxy}}$