Given arctan y/x = ln sqrt(x²+y²), verify that x (dy-dx)

Given $\arctan\dfrac{y}{x} = \ln\sqrt{x^{2}+y^{2}}$, verify that $x\bigl(dy-dx\bigr) = y\bigl(dx+dy\bigr)$.

Step 1: Translate the target into a statement about $y'$

Expand both sides and gather the $dy$ terms on the left, the $dx$ terms on the right: \[x\,dy-x\,dx = y\,dx+y\,dy \;\Longleftrightarrow\; \bigl(x-y\bigr)dy = \bigl(x+y\bigr)dx \;\Longleftrightarrow\; y' = \frac{x+y}{x-y}.\]

So the identity to be proved is exactly the slope formula for this curve, written without denominators.

Step 2: Rewrite the given relation

The logarithm of a square root is half the logarithm: \[\ln\sqrt{x^{2}+y^{2}} = \frac12\ln\bigl(x^{2}+y^{2}\bigr), \qquad\text{so}\qquad \arctan\frac{y}{x} = \frac12\ln\bigl(x^{2}+y^{2}\bigr).\]

Step 3: Differentiate the left-hand side

\[ \frac{d}{dx}\arctan\frac{y}{x} = \frac{1}{1+\dfrac{y^{2}}{x^{2}}}\cdot\frac{y'x-y}{x^{2}} = \frac{y'x-y}{x^{2}+y^{2}}. \]

Step 4: Differentiate the right-hand side

\[ \frac{d}{dx}\left[\frac12\ln\bigl(x^{2}+y^{2}\bigr)\right] = \frac{x+yy'}{x^{2}+y^{2}}. \]

Step 5: Cancel and solve

The common denominator $x^{2}+y^{2}$ is nonzero on the curve, so it cancels: \[y'x-y = x+yy' \;\Longrightarrow\; y'\bigl(x-y\bigr) = x+y \;\Longrightarrow\; y' = \frac{x+y}{x-y}.\]

Step 6: Return to differentials

Multiplying $\bigl(x-y\bigr)dy = \bigl(x+y\bigr)dx$ out and regrouping: \[x\,dy-y\,dy = x\,dx+y\,dx \;\Longrightarrow\; x\,dy-x\,dx = y\,dx+y\,dy \;\Longrightarrow\; x\bigl(dy-dx\bigr) = y\bigl(dx+dy\bigr).\]

Step 7: Verification

\[\text{In polar form the relation is } \varphi = \ln r, \text{ i.e. } r = e^{\varphi}\]

\[\varphi = 0.1:\ (x,y) \approx (1.0996497,\ 0.1103330), \qquad \frac{x+y}{x-y} \approx 1.2230489\]

\[x\bigl(y'-1\bigr) = 1.0996497\cdot 0.2230489 \approx 0.2452756; \qquad y\bigl(1+y'\bigr) = 0.1103330\cdot 2.2230489 \approx 0.2452756 \;\;\checkmark\]