Mathematical Immaturity

14.19 Exercises

• Prove that if a homogeneous first-order differential equation of the form $y' = f(x, y)$ is rewritten in polar coordinates, it reduces to a separable equation. Use this method to solve $y' = (y - x)/(y + x).$
• To rewrite the equation in polar coordinates, we must first find $r$ and $\theta$ in terms of $x$ and $y:$ \begin{align*} \\ r &= \sqrt{x^2 + y^2}, \quad \theta = \arctan\frac{y}{x} \end{align*} Then, we can rewrite $(x, y)$ in polar coordinates with $x = r\cos\theta$ and $y = r\sin\theta,$ where $r$ is a function of $\theta,$ giving us: \begin{align*} \\ \frac{dy}{d\theta} &= \frac{dr}{d\theta}\sin\theta + r\cos\theta \\ \\ \frac{dx}{d\theta} &= \frac{dr}{d\theta}\cos\theta - r\sin\theta \end{align*} Then, we use the chain rule to show that \begin{align*} \\ \frac{dy}{d\theta} &= \frac{dy}{dx}\frac{dx}{d\theta} \\ \\ &= y'\frac{dx}{d\theta} \end{align*} In other words, for $\frac{dx}{d\theta} \neq 0,$ the equation for $y'$ becomes: \begin{align*} \\ y' &= \frac{dy/d\theta}{dx/d\theta} \\ \\ &= \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} \\ \\ &= f(r\cos\theta,r\sin\theta) \end{align*} Rearranging terms we find that \begin{align*} \\ y'\left(r'\cos\theta - r\sin\theta\right) &= r'\sin\theta + r\cos\theta \\ \\ r'\left(y'\cos\theta - \sin\theta\right) &= r\left(y'\sin\theta + \cos\theta\right) \\ \\ \frac{r'}{r} &= \frac{y'\sin\theta + \cos\theta}{y'\cos\theta - \sin\theta} \end{align*} If we denote the right-hand side as $Q(\theta),$ then we can see that this is a separable equation in terms of $r$ and $Q.$ Now, to solve $y' = (y - x)/(y + x),$ we write $y'$ in terms of polar coordinates: \begin{align*} \\ y' &= \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} \end{align*} Setting $x = r\cos\theta$ and $y = r\sin\theta,$ the given equation for $y'$ becomes: \begin{align*} \\ y' &= \frac{y - x}{y + x} \\ \\ &= \frac{r\sin\theta - r \cos\theta}{r \sin\theta + r \cos\theta} \\ \\ &= \frac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta} \end{align*} Now, we can plug this into our separable equation in $r:$ \begin{align*} \\ \frac{r'}{r} &= \frac{\frac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta}\sin\theta + \cos\theta}{\frac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta}\cos\theta - \sin\theta} \\ \\ &= \frac{\left(\sin\theta - \cos\theta\right)\sin\theta + \left(\sin\theta + \cos\theta\right)\cos\theta}{\left(\sin\theta - \cos\theta\right)\cos\theta - \left(\sin\theta + \cos\theta\right)\sin\theta} \\ \\ &= -1 \end{align*} Integrating both sides, we get $\log r +\theta = C.$ But since $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan \frac{y}{x},$ we reach the conclusion: \begin{align*} \\ \\ \log \sqrt{x^2 + y^2} + \arctan \left(\frac{y}{x}\right) &= C \quad \blacksquare \\ \\ \end{align*}