Example 10: Homogeneous: $(x - y)\frac{dy}{dx} = x + 2y$

Basics you need

Solution

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

   For $x \ne y$⁠, divide by $x - y$⁠:

   $$\frac{dy}{dx} = \frac{x + 2y}{x - y} = F(x, y).\quad\ldots(1)$$
   The homogeneity test is stated for $\frac{dy}{dx} = F(x, y)$⁠. Dividing loses nothing: on the line $y = x$ (with $x \ne 0$⁠) the given equation would read $0 = x + 2x = 3x \ne 0$⁠, so no solution curve $y(x)$ can pass through such a point anyway.
   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: $F(\lambda x, \lambda y) = F(x, y)$

   Replace $x$ by $\lambda x$ and $y$ by $\lambda y$⁠, for any $\lambda \ne 0$⁠:

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{\lambda x + 2\lambda y}{\lambda x - \lambda y} = \frac{\lambda(x + 2y)}{\lambda(x - y)} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

   So $F$ is a homogeneous function of degree zero, and (1) is a homogeneous differential equation. Equivalently, dividing top and bottom by $x$⁠, $F(x, y) = \frac{1 + 2\frac{y}{x}}{1 - \frac{y}{x}}$⁠, a function of $\frac{y}{x}$ alone.

   The $\lambda$ cancels because top and bottom both have degree 1. "Degree zero" is exactly the condition for the equation to be homogeneous. (The book writes the result as $\lambda^0\cdot f(x, y)$ with a small $f$⁠; it means the same function $F$⁠.)
   Section 9.4.2: definition of a homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$⁠, where $v$ is a function of $x$⁠. By the product rule,

   $$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$

   Substituting in (1), and dividing top and bottom by $x$⁠:

   $$v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx} = \frac{1 + 2v}{1 - v}.$$
   Because $F$ depends only on $\frac{y}{x} = v$⁠, the right side becomes a function of $v$ only. The left side has $x\frac{dv}{dx}$⁠, which is what will let us separate $v$ from $x$⁠. Here $x \ne 0$⁠, since $v = \frac{y}{x}$⁠.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

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

   $$\begin{aligned} x\frac{dv}{dx} &= \frac{1 + 2v}{1 - v} - v \\ &= \frac{1 + 2v - v(1 - v)}{1 - v} \\ &= \frac{v^2 + v + 1}{1 - v}. \end{aligned}$$
   Subtract $v$ by writing it over the common denominator $1 - v$⁠: $1 + 2v - v + v^2 = v^2 + v + 1$⁠. Now the right side is a function of $v$ only, and the left side is $x\frac{dv}{dx}$⁠: the equation is separable.
   Section 9.4.2, equation (4): $x\frac{dv}{dx} = g(v) - v$⁠. Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\begin{aligned} \frac{1 - v}{v^2 + v + 1}\,dv &= \frac{dx}{x}, \quad\text{that is} \\ \frac{v - 1}{v^2 + v + 1}\,dv &= -\frac{dx}{x}. \end{aligned}$$
   Multiply by $\frac{1 - v}{v^2 + v + 1}$ and by $\frac{dx}{x}$⁠. This is allowed: $v^2 + v + 1 = \left(v + \frac12\right)^2 + \frac34 > 0$ for every $v$⁠, and $x \ne 0$⁠. Writing $v - 1$ on top (and a minus on the right) matches the book's form.
   Section 9.4.2, equation (5); separating as in Section 9.4.1. Basics: dy = y′ dx

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

   Split the numerator using the derivative of the denominator, $\frac{d}{dv}(v^2 + v + 1) = 2v + 1$⁠:

   $$v - 1 = \tfrac12(2v + 1) - \tfrac32.$$

   So

   $$\begin{aligned} &\int\frac{v - 1}{v^2 + v + 1}\,dv \\ &= \frac12\int\frac{2v + 1}{v^2 + v + 1}\,dv \\ &\quad - \frac32\int\frac{dv}{\left(v + \frac12\right)^2 + \left(\frac{\sqrt3}{2}\right)^2}. \end{aligned}$$

   First part: put $t = v^2 + v + 1$⁠, $dt = (2v + 1)\,dv$⁠: it is $\frac12\log|t| = \frac12\log(v^2 + v + 1)$⁠.

   Second part: with $a = \frac{\sqrt3}{2}$⁠, Section 7.4 formula (3) gives

   $$\begin{aligned} \int\frac{dv}{\left(v + \frac12\right)^2 + a^2} &= \frac{1}{a}\tan^{-1}\frac{v + \frac12}{a} \\ &= \frac{2}{\sqrt3}\tan^{-1}\frac{2v + 1}{\sqrt3}. \end{aligned}$$

   The $x$ side is $-\int\frac{dx}{x} = -\log|x|$⁠. Hence, with one constant $C_1$⁠,

   $$\begin{aligned} &\frac12\log(v^2 + v + 1) - \sqrt3\tan^{-1}\frac{2v + 1}{\sqrt3} \\ &= -\log|x| + C_1.\quad\ldots(2) \end{aligned}$$

   Check by differentiating the left side of (2) with respect to $v$⁠:

   $$\begin{aligned} &\frac{2v + 1}{2(v^2 + v + 1)} - \sqrt3\cdot\frac{\frac{2}{\sqrt3}}{1 + \frac{(2v + 1)^2}{3}} \\ &= \frac{2v + 1}{2(v^2 + v + 1)} - \frac{6}{4(v^2 + v + 1)} \\ &= \frac{v - 1}{v^2 + v + 1}. \end{aligned}$$
   An integrand $\frac{\text{linear}}{\text{quadratic}}$ is split into (a multiple of the derivative of the bottom) plus (a constant): the first piece gives a log, the second a $\tan^{-1}$ after completing the square. The coefficient is $\frac32\cdot\frac{2}{\sqrt3} = \sqrt3$⁠. In the check, $3 + (2v + 1)^2 = 4(v^2 + v + 1)$⁠.
   Section 7.4, formula (3), and $\int\frac{f'}{f} = \log|f|$ by substitution. Basics: every method ends with an integral

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

   Move $-\log|x|$ to the left, using $\log|x| = \frac12\log x^2$ (this answers the book's "(Why?)": $x^2 = |x|^2$⁠):

   $$\begin{aligned} &\frac12\log(v^2 + v + 1) + \frac12\log x^2 \\ &= \sqrt3\tan^{-1}\frac{2v + 1}{\sqrt3} + C_1. \end{aligned}$$

   Combine the logs, $\frac12\log\big[(v^2 + v + 1)x^2\big]$⁠, and put $v = \frac{y}{x}$⁠:

   $$\left(\frac{y^2}{x^2} + \frac{y}{x} + 1\right)x^2 = y^2 + xy + x^2,$$ $$\frac{2v + 1}{\sqrt3} = \frac{2y + x}{\sqrt3x}.$$

   Multiply by 2 and write $C = 2C_1$⁠:

   $$\begin{aligned} &\log|x^2 + xy + y^2| \\ &= 2\sqrt3\tan^{-1}\frac{x + 2y}{\sqrt3x} + C. \end{aligned}$$
   Both log arguments are positive ($x^2 > 0$ and $v^2 + v + 1 > 0$⁠), so they may be combined. Since $x^2 + xy + y^2 > 0$ away from the origin, the absolute value bars in the book's answer change nothing. $C = 2C_1$ is again an arbitrary constant, the one constant of a first order equation.
   $\log a + \log b = \log ab$ ($a, b > 0$⁠). Basics: one constant for order 1

8. Check: differentiate the answer back

   Write $Q = x^2 + xy + y^2$ and $u = \frac{x + 2y}{\sqrt3x}$⁠. Then $1 + u^2 = \frac{4Q}{3x^2}$ and $\frac{du}{dx} = \frac{2(xy' - y)}{\sqrt3x^2}$⁠. Differentiating the answer with respect to $x$⁠:

   $$\begin{aligned} &\frac{2x + y + (x + 2y)y'}{Q} \\ &= 2\sqrt3\cdot\frac{3x^2}{4Q}\cdot\frac{2(xy' - y)}{\sqrt3x^2} \\ &= \frac{3(xy' - y)}{Q}, \end{aligned}$$

   so, multiplying by $Q$⁠,

   $$\begin{aligned} 2x + y + (x + 2y)y' &= 3xy' - 3y \\ (2y - 2x)\,y' &= -(2x + 4y) \\ y' &= \frac{x + 2y}{x - y}. \end{aligned}$$
   We recover the given equation for every value of $C$⁠, so the relation is the general solution. The picture shows homogeneity at work: $F(\lambda x, \lambda y) = F(x, y)$ means the slope is the same at all points of a ray through the origin, and a curve reflected in the origin is again a solution.
   Implicit differentiation, chain rule, $\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^2}$ (Chapter 5). Basics: a rule for slopes

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