Show that the function y defined implicitly by arctan y/x

Show that the function $y$ defined implicitly by $\arctan\dfrac{y}{x} = \dfrac12\ln\bigl(x^{2}+y^{2}\bigr)$ satisfies the equation $y'(x-y) = x+y$.

Step 1: Differentiate the left-hand side

With $u = \dfrac{y}{x}$, the quotient rule gives $u' = \dfrac{y'x-y}{x^{2}}$, and $\bigl(\arctan u\bigr)' = \dfrac{u'}{1+u^{2}}$: \[\frac{d}{dx}\arctan\frac{y}{x} = \frac{1}{1+\dfrac{y^{2}}{x^{2}}}\cdot\frac{y'x-y}{x^{2}} = \frac{x^{2}}{x^{2}+y^{2}}\cdot\frac{y'x-y}{x^{2}} = \frac{y'x-y}{x^{2}+y^{2}}.\]

The factor $x^{2}$ cancels completely — this is what makes the result so clean.

Step 2: Differentiate the right-hand side

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

Step 3: Equate and cancel the denominator

Both sides carry the same denominator $x^{2}+y^{2}$, which is nonzero on the curve, so it cancels: $\frac{y'x-y}{x^{2}+y^{2}} = \frac{x+yy'}{x^{2}+y^{2}} \;\Longrightarrow\; y'x-y = x+yy'$.

Step 4: Collect the $y'$ terms

\[ y'x - yy' = x+y \;\Longrightarrow\; y'(x-y) = x+y, \]

which is the required equation. Hence also

\[ y' = \frac{x+y}{x-y} \qquad (x\neq y). \]

Step 5: Verification

In polar coordinates $x = r\cos\theta,\ y = r\sin\theta$, the relation reads $\theta = \ln r$, i.e. $r = e^{\theta}$ — a logarithmic spiral. Testing a point on it: $\theta = 0.1:\quad r = e^{0.1},\ (x,y) \approx (1.0996497,\ 0.1103330)$.

\[\arctan\frac{y}{x} = 0.100000 = \frac12\ln\bigl(x^{2}+y^{2}\bigr) \;\;\checkmark\]

\[y' = \frac{x+y}{x-y} \approx 1.2230489; \qquad \text{numerically } \frac{\Delta y}{\Delta x} \approx 1.2230489 \;\;\checkmark\]

\[\theta = 0:\quad (x,y) = (1,0), \qquad y' = \frac{1+0}{1-0} = 1 \;\;\checkmark\]