Question

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

Answer 1

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

first of all, resolve xdx + ydy
$xdx+ydy=\frac{1}{2}(2xdx+2ydy)\\\\=\frac{1}{2}d(x^2+y^2)$

you know, $d(tan^{-1}P)=\frac{1}{1+P^2}dP$
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 $\frac{x^2}{2}+\frac{y^2}{2}=a^2tan^{-1}\frac{y}{x}+C$

Answer 2

Answer:

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

Step-by-step explanation:

Given expression is

$xdx + ydy = a^{2} (\frac{xdy - ydx}{x^{2} + y^{2} } )$ .... (A)

As, we need to find in the solution of above expression, so, let us take left hand side first, which can be written as follows

$xdx + ydy = \frac{1}{2} (2xdx + 2ydy)$

so, we can write the above equation as

$xdx + ydy = \frac{1}{2} d(x^{2} + y^{2})$ .....(1)

As we know already

$d(tan^{-1}A)= \frac{1}{1 + A^{2}} . dA$

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

So

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

multiply both sides of above equation by $a^{2}$

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

$a^{2}d(tan^{-1}(\frac{y}{x}))$ is equation to right hand side of equation (A).

Now substituting left hand side of equation A with $\frac{1}{2} d(x^{2} + y^{2})$ from equation (1) and right hand side with $a^{2}d(tan^{-1}(\frac{y}{x}))$ of equation 2, we get

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

The differential will be cancelled on both sides and we will get following equation

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

This is our solution