Exercise 9.4 · Q3: Homogeneous: $(x - y)\,dy - (x + y)\,dx = 0$

Basics you need

Solution

1. Write the equation as $\frac{dy}{dx} = F(x, y)$

   Move $(x + y)\,dx$ across and divide by $(x - y)\,dx$⁠, for $x \ne y$⁠:

   $$\frac{dy}{dx} = \frac{x + y}{x - y} = F(x, y).\quad\ldots(1)$$
   Dividing by $x - y$ loses no solution: at a point of the line $y = x$ (with $x \ne 0$⁠) the given equation reads $0\cdot dy = 2x\,dx$⁠, which forces $dx = 0$⁠, a vertical tangent. So no solution $y(x)$ with a finite slope passes through such a point. We also need $x \ne 0$ for $v = \frac{y}{x}$⁠.
   Section 9.4.2: a homogeneous equation has the form $\frac{dy}{dx} = F(x, y)$⁠. Lesson: homogeneous equations

2. Show that it is homogeneous: $F(\lambda x, \lambda y) = F(x, y)$

   For any $\lambda \ne 0$⁠:

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{\lambda x + \lambda y}{\lambda x - \lambda y} = \frac{\lambda(x + y)}{\lambda(x - y)} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

   So $F$ is homogeneous of degree zero, and (1) is a homogeneous differential equation.

   Top and bottom have degree 1, so $\lambda$ cancels. Dividing top and bottom by $x$ gives $F = \frac{1 + y/x}{1 - y/x}$⁠, a function of $\frac{y}{x}$ alone.
   Section 9.4.2: definition of a homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$⁠. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠, and (1) becomes

   $$v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}.$$
   Divide top and bottom by $x \ne 0$⁠: the right side becomes a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = \ldots$

   $$\begin{aligned} x\frac{dv}{dx} &= \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} \\ &= \frac{1 + v^2}{1 - v}. \end{aligned}$$
   Write $v$ over the common denominator $1 - v$⁠: $v(1 - v) = v - v^2$⁠, and $1 + v - (v - v^2) = 1 + v^2$⁠.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\frac{1 - v}{1 + v^2}\,dv = \frac{dx}{x}.$$
   Multiply by $\frac{1 - v}{1 + v^2}$ and divide by $x$⁠. Since $1 + v^2 \ge 1$ is never zero, this division sets no solution aside; $v \ne 1$ and $x \ne 0$ were assumed already.
   Section 9.4.2, equation (5). Basics: dy = y′ dx

6. Integrate: a $\tan^{-1}$ part and a log part

   Split the fraction and integrate each piece:

   $$\begin{aligned} &\int\frac{1 - v}{1 + v^2}\,dv \\ &= \int\frac{dv}{1 + v^2} - \frac12\int\frac{2v\,dv}{1 + v^2} \\ &= \tan^{-1}v - \frac12\log(1 + v^2). \end{aligned}$$

   With $\int\frac{dx}{x} = \log|x|$ and one constant $C$⁠:

   $$\begin{aligned} &\tan^{-1}v - \frac12\log(1 + v^2) \\ &= \log|x| + C.\quad\ldots(2) \end{aligned}$$

   Check by differentiating the left side with respect to $v$⁠:

   $$\frac{1}{1 + v^2} - \frac12\cdot\frac{2v}{1 + v^2} = \frac{1 - v}{1 + v^2}.$$
   The top $2v$ is the derivative of $1 + v^2$⁠, so that piece is a log (substitution $t = 1 + v^2$⁠); $1 + v^2 > 0$⁠, so no bars are needed there. The check returns the integrand.
   Section 7.2, formulas (x) and (xii); substitution, Section 7.3.1. Basics: every method ends with an integral

7. Put $v = \frac{y}{x}$ back and tidy up

   Move the log to the right and use $\log|x| = \frac12\log x^2$⁠:

   $$\begin{aligned} \tan^{-1}\frac{y}{x} &= \frac12\log\left(1 + \frac{y^2}{x^2}\right) \\ &\quad + \frac12\log x^2 + C \\ &= \frac12\log\left(\frac{x^2 + y^2}{x^2}\cdot x^2\right) + C \\ &= \frac12\log(x^2 + y^2) + C. \end{aligned}$$
   Both log arguments are positive, so they combine into one log. The answer holds for $x \ne 0$⁠, $y \ne x$⁠; $C$ is the one arbitrary constant.
   $\log a + \log b = \log ab$ for $a, b > 0$⁠. Basics: one constant for order 1

8. Check, and see the curves

   Differentiate the answer with respect to $x$ (implicitly):

   $$\begin{aligned} &\frac{d}{dx}\tan^{-1}\frac{y}{x} \\ &= \frac{1}{1 + \frac{y^2}{x^2}}\cdot\frac{xy' - y}{x^2} = \frac{xy' - y}{x^2 + y^2}, \end{aligned}$$ $$\frac{d}{dx}\,\frac12\log(x^2 + y^2) = \frac{x + yy'}{x^2 + y^2}.$$

   Setting them equal: $xy' - y = x + yy'$⁠, so $(x - y)\,y' = x + y$⁠, the given equation.

   In polar coordinates $x = r\cos\theta$⁠, $y = r\sin\theta$ with $x > 0$⁠, we have $\tan^{-1}\frac{y}{x} = \theta$ and $\frac12\log(x^2 + y^2) = \log r$⁠, so the answer reads $\theta = \log r + C$⁠, that is $r = e^{\theta - C}$⁠: a spiral that winds out as $\theta$ grows.

   The relation gives back the equation, so it is the general solution. On the left half-plane $x < 0$⁠, $\tan^{-1}\frac{y}{x} = \theta - \pi$⁠, so a whole spiral is described by values of $C$ that differ by $\pi$ on the two sides of the $y$-axis; each piece is a solution on its own half-plane.
   Implicit differentiation; $\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^2}$ (Chapter 5). Basics: a rule for slopes

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