Can I equate the solutions $\arcsin\frac{x^2+y^2-a^2/2}{a^2/2}=2\arctan\frac{y}{x}+C$ to $\arcsin\frac{\sqrt{x^2+y^2}}{a}=\arctan\frac{y}{x}+C$

Question

$$ \frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}} $$

This is asked many times and attmpted to solve particularly at solve$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ by @Simon S by seting $u=x^2+y^2$ and $v=\dfrac{y}{x}$ $$ LHS=\frac{du}{2x^2dv}=\frac{du}{2dv}.\frac{1+v^2}{u}\\ RHS=\sqrt{\frac{a^2-u}{u}}\\ \frac{du}{2dv}.\frac{1+v^2}{u}=\sqrt{\frac{a^2-u}{u}}\implies\boxed{\int\frac{du}{\sqrt{u(a^2-u)}}=\int\frac{du}{\sqrt{-[u^2-a^2u}]}=2\int\frac{dv}{1+v^2}}\\ \int\frac{du}{\sqrt{-[u^2-2\frac{a^2}{2}u+\frac{a^4}{4}-\frac{a^4}{4}]}}=\int\frac{du}{\sqrt{\Big(\dfrac{a^2}{2}\Big)^2-\Big(u-\dfrac{a^2}{2}\Big)^2}}=2\int\frac{dv}{1+v^2}\\ \implies \sin^{-1}\bigg[\frac{u-\dfrac{a^2}{2}}{\dfrac{a^2}{2}}\bigg]=2\tan^{-1}v+C\\ \color{red}{\sin^{-1}\bigg[\frac{x^2+y^2-\dfrac{a^2}{2}}{\dfrac{a^2}{2}}\bigg]=2\tan^{-1}\frac{y}{x}+C}\\ $$

But my reference and the post that I mentioned give the solution $\sqrt{x^2+y^2}=a\sin\Big[C+\tan^{-1}\frac{y}{x}\Big]\implies\color{red}{\sin^{-1}\frac{\sqrt{x^2+y^2}}{a}=\tan^{-1}\frac{y}{x}+C}$, so are both the same or am I missing something here ?

Note: This isn't about exactly how to solve the differential equation, which is what is asked in the mentioned post(already specified by me in the first place), I am having trouble equating the form of solution given there, kindly look into it before marking it duplicate.

Answer 1

Set $\sqrt{u}=a\sin\theta\implies u=a^2\sin^2\theta\implies du=2a^2\sin\theta\cos\theta.d\theta$ $$ \int\frac{2a^2\sin\theta\cos\theta.d\theta}{a\sin\theta.a\cos\theta}=2\int d\theta=2\int\frac{dv}{1+v^2}\\ \theta=\sin^{-1}\Big[\frac{\sqrt{u}}{a}\Big]=\tan^{-1}\big(v\big)+C\\\\ \sin^{-1}\Big[\frac{\sqrt{x^2+y^2}}{a}\Big]=\tan^{-1}\big(\frac{y}{x}\big)+C\\ \boxed{\sqrt{x^2+y^2}=a\sin\Big[\tan^{-1}\big(\frac{y}{x}\big)+C\Big]} $$

Answer 2

$$ \sin^{-1}\frac{\sqrt{x^2+y^2}}{a}=\tan^{-1}\frac{y}{x}+C \tag1$$

translates, using polar coordinate into

$$ r= a\sin (\theta + \tan^{-1}{}C)\tag2$$

Geometrically these are all circles passing through origin with diameter $a.$

Second integrand (take 2C as constant)

$${\sin^{-1}\bigg[\frac{x^2+y^2-\dfrac{a^2}{2}}{\dfrac{a^2}{2}}\bigg]=2\tan^{-1}\frac{y}{x}+2C} $$

sine both sides

$$\dfrac{ 2r^2}{a^2}=1+ \sin (2 \theta+ 2C) \tag3$$

$$ \frac{ r^2}{a^2}= \large[ \frac{\sin(\theta+C)}{\sqrt 2}+\frac{\cos(\theta+C)}{\sqrt 2}\large]^2 $$

$$ =\large[\sin({ \theta +C+\frac{\pi}{4}}) \large]^2\tag4$$

$$ r= a \sin (\theta+C_1)\tag5 $$

which is the same solution except for an arbitrary constant of integration represented in a circle rotated by a different angle $C.$ It is just a trig exercise to recognize the constant in different form.