Question

xdx+ydy=a^2(xdy-ydx)/x^2+y^2

Answer 1

Answer:

Step-by-step explanation:

I think that your question may be

$xdx+ydy={a}^2(xdy+ydx)/({x}^2+{y}^2)$

$({x}^2+{y}^2)(xdx+ydy)={a}^2(xdy+ydx)..(1)$

$Take, \\u={x}^2+{y}^2\\\frac{du}{dx}=2x+2y\frac{dy}{dx}\\\frac{du}{dx}=2(x+y\frac{dy}{dx})\\\frac{du}{2}=xdx+ydy$

$v=xy\\\frac{dv}{dx}=x\frac{dy}{dx}+y.1\\dv=xdy+ydx\\Now (1) becomes$

$u\frac{du}{2}={a}^2dv\\Integrating\\\frac{1}{2}\int\:u\:du={a}^2\int\:dv$

$\frac{{u}^2}{2}={a}^2v+c\\\frac{{({x}^2+{y}^2)}^2}{2}={a}^2xy+c$

Answer 2

it is given that, $xdx+ydy=a^2\frac{xdy-ydx}{x^2+y^2}$

first of all, resolve xdx + ydy 
xdx + ydy = 1/2(2xdx + 2ydy)
= 1/2d(x² + y²)

you know, $d(tan^{-1}P)=\frac{1}{1+P^2}$

so, $d\left(tan^{-1}\frac{y}{x}\right)=\frac{x^2}{x^2+y^2}\frac{xdy-ydx}{x^2}$

= $d\left(tan^{-1}\frac{y}{x}\right)=\frac{xdy-ydx}{x^2+y^2}$

hence, $xdx+ydy=a^2\frac{xdy-ydx}{x^2+y^2}$ is converted into 
$\frac{1}{2}d(x^2+y^2)=a^2d\left(tan^{-1}\frac{y}{x}\right)$

or, $d\left(\frac{x^2}{2}+\frac{y^2}{2}\right)=d\left(a^2tan^{-1}\frac{y}{x}\right)$

or, $\frac{x^2}{2}+\frac{y^2}{2}=a^2tan^{-1}\frac{y}{x}+C$

hence, solution is $\frac{x^2}{2}+\frac{y^2}{2}=a^2tan^{-1}\frac{y}{x}+C$