How can I prove that: $x(dy-dx)=y(dy+dx)$?

Question

given $y=f(x)$ so that:

$$ \arctan\bigg( \frac{y}{x} \bigg) = \ln\big( \sqrt{x^2 + y^2} \big) $$

How can I prove that: $x(dy-dx)=y(dy+dx)$ ?

Answer 1

By implicit differentiation you get $$\frac{d}{dx}\arctan(y/x)=\frac{1}{1+(y/x)^2}\frac{d}{dx}(y/x)=\frac{1}{1+(y/x)^2}\left(-y/x^2+ y'/x\right)=\frac{-y+xy'}{x^2+y^2}$$ and $$\frac{d}{dx}\ln\sqrt{x^2+y^2}=\frac{1}{\sqrt{x^2+y^2}}\frac{d}{dx}\sqrt{x^2+y^2}=\frac{1}{\sqrt{x^2+y^2}}\left(\frac{2x+2yy'}{2\sqrt{x^2+y^2}}\right)=\frac{x+yy'}{x^2+y^2}$$ Thus, $$\frac{-y+xy'}{x^2+y^2}=\frac{x+yy'}{x^2+y^2}$$ or $$-y+x\frac{dy}{dx}=x+y\frac{dy}{dx}$$ $$-ydx+xdy=xdx+ydy$$ $$x(dy-dx)=y(dx+dy)$$

Answer 2

Note that

$$x(dy-dx)=y(dy+dx)\implies(x-y)dy=(x+y)dx\implies\frac{dy}{dx}=\frac{x+y}{x-y}$$

In polar coordinates

$$\arctan\bigg( \frac{y}{x} \bigg) = \ln\big( \sqrt{x^2 + y^2} \big)\implies\theta=\ln\rho\implies d\theta=\frac{d\rho}{\rho}$$

and

$$dx=-\rho\sin\theta \, d\theta+\cos\theta \, d\rho$$

$$dy=\rho\cos\theta \, d\theta+\sin\theta \, d\rho$$

then

$$\frac{dy}{dx}=\frac{\rho\cos\theta \, d\theta+\sin\theta \,d\rho}{-\rho\sin\theta \,d\theta+\cos\theta \,d\rho}=\frac{\cos\theta +\sin\theta}{-\sin\theta +\cos\theta }=\frac{x+y}{x-y}$$

Answer 3

We are given $$\arctan\bigg( \frac{y}{x} \bigg) = \ln\big( \sqrt{x^2 + y^2} \big) =(1/2)ln(x^2+y^2)$$

Upon differentiation of both sides we get:$$ \frac {d(y/x)}{1+(y/x)^2}= (1/2)\frac {d(x^2+y^2)}{x^2+y^2}\text { (*)} $$Note that $$d(y/x) = \frac {xdy-ydx}{x^2} $$ and $$d(x^2+y^2) = 2xdx+2ydy$$ Upon substitution of these results in $ (*)$ and a little algebra we get the $$x(dy-dx)=y(dy+dx)$$