Exercise 9.4 · Q4: Homogeneous: $(x^2 - y^2)\,dx + 2xy\,dy = 0$

Basics you need

Solution

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

   For $x \ne 0$⁠, $y \ne 0$⁠, move $(x^2 - y^2)\,dx$ across and divide by $2xy\,dx$⁠:

   $$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy} = F(x, y).\quad\ldots(1)$$
   We need $x \ne 0$ for $v = \frac{y}{x}$⁠. Dividing by $y$ loses no solution: at a point $(x_0, 0)$ with $x_0 \ne 0$ the given equation reads $x_0^2\,dx = 0$⁠, which forces $dx = 0$⁠, a vertical tangent. So a solution curve can only cross the $x$-axis vertically, and no solution $y(x)$ with a finite slope passes through it. (The $y$-axis $x = 0$ also satisfies the equation written with differentials, since there $dx = 0$ and $2xy = 0$⁠; it is not a graph $y(x)$⁠, and it lies outside the domain $x \ne 0$ of $v = \frac{y}{x}$⁠.)
   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 $\lambda \ne 0$⁠:

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

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

   Top and bottom both have degree 2, so $\lambda^2$ cancels. Dividing top and bottom by $x^2$ gives $F = \frac{(y/x)^2 - 1}{2(y/x)}$⁠, 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

   $$v + x\frac{dv}{dx} = \frac{v^2x^2 - x^2}{2vx^2} = \frac{v^2 - 1}{2v}.$$
   Cancel $x^2 \ne 0$⁠: the right side is a function of $v$ only. Here $v \ne 0$⁠, since $y \ne 0$⁠.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

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

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

5. Separate $v$ and $x$

   $$\frac{2v}{1 + v^2}\,dv = -\frac{dx}{x}.$$
   Multiply by $\frac{2v}{1 + v^2}$ and divide by $x$⁠. Since $1 + v^2 \ge 1$⁠, dividing by it loses nothing; $x \ne 0$ and $v \ne 0$ were assumed already.
   Section 9.4.2, equation (5). Basics: dy = y′ dx

6. Integrate, with the constant as $\log|C|$

   The top $2v$ is the derivative of $1 + v^2$⁠, so (put $t = 1 + v^2$⁠, $dt = 2v\,dv$⁠)

   $$\int\frac{2v}{1 + v^2}\,dv = \log(1 + v^2).$$

   Hence, writing the one arbitrary constant as $\log|C|$ ($C \ne 0$⁠),

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

   Check: $\frac{d}{dv}\log(1 + v^2) = \frac{2v}{1 + v^2}$⁠, the integrand.

   $\log|C|$ runs over all real numbers as $C$ runs over the nonzero reals, so it is just as arbitrary as any other constant; this form lets the logs combine. No bars are needed on $1 + v^2 > 0$⁠.
   $\int\frac{f'}{f} = \log|f|$ by substitution (Section 7.3.1); Section 7.2, formula (xii). Basics: every method ends with an integral

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

   Combine the logs: $\log\left[(1 + v^2)|x|\right] = \log|C|$⁠, so $(1 + v^2)|x| = |C|$⁠. With $v = \frac{y}{x}$⁠:

   $$\begin{aligned} \frac{x^2 + y^2}{x^2}\cdot|x| &= |C|,\ \text{that is} \\ \frac{x^2 + y^2}{|x|} &= |C|. \end{aligned}$$

   Removing the bars, $\frac{x^2 + y^2}{x} = \pm C$⁠; naming the constant $C$ again,

   $$x^2 + y^2 = Cx.$$
   Since $x^2 + y^2 > 0$⁠, $C$ has the sign of $x$⁠: curves with $C > 0$ lie in $x > 0$⁠, curves with $C < 0$ in $x < 0$⁠. Completing the square, $\left(x - \frac{C}{2}\right)^2 + y^2 = \frac{C^2}{4}$⁠: circles through the origin with centres on the $x$-axis, touching the $y$-axis at $O$⁠. Each circle gives two solutions $y(x)$⁠, its upper and lower halves, which meet the $x$-axis vertically, as Step 1 predicted.
   $\log a + \log b = \log ab$⁠; $\log A = \log B \Rightarrow A = B$⁠. Basics: one constant for order 1

8. Check, and see the circles

   Differentiate $x^2 + y^2 = Cx$ with respect to $x$⁠: $2x + 2yy' = C$⁠. Put $C = \frac{x^2 + y^2}{x}$ and multiply by $x$⁠:

   $$\begin{aligned} 2x^2 + 2xyy' &= x^2 + y^2 \\ 2xy\,y' &= y^2 - x^2, \end{aligned}$$

   which is $(x^2 - y^2)\,dx + 2xy\,dy = 0$⁠.

   We recover the given equation for every $C$⁠, so $x^2 + y^2 = Cx$ is the general solution. With equal scales the circles look round and the slope marks lie along them.
   Implicit differentiation (Chapter 5). Basics: a rule for slopes

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