Example 13: Curves with slope $\frac{x^2+y^2}{2xy}$ are $x^2 - y^2 = cx$

Basics you need

Solution

1. Turn the slope condition into an equation

   $$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy} = F(x, y).\quad\ldots(1)$$
   The slope of the tangent at $(x, y)$ is $\frac{dy}{dx}$⁠. The given expression needs $xy \ne 0$⁠, so every curve of the family stays off both axes.
   The slope of the tangent is $\frac{dy}{dx}$⁠, as used in Section 9.4.1, Example 8. Basics: a rule for slopes

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

   For any $\lambda \ne 0$⁠:

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

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

   Top and bottom both have degree 2, so the same $\lambda^2$ comes out of each and cancels. The book says this is "clear"; the line above is the actual reason. Dividing top and bottom by $x^2$ shows the same thing: $F = \frac{1 + (y/x)^2}{2(y/x)}$⁠, a function of $\frac{y}{x}$ alone.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$ and simplify

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

   $$\begin{aligned} v + x\frac{dv}{dx} &= \frac{x^2 + v^2x^2}{2vx^2} = \frac{1 + v^2}{2v} \\ x\frac{dv}{dx} &= \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} \\ &= \frac{1 - v^2}{2v}. \end{aligned}$$
   The $x^2$ cancels because $F$ has degree zero. Subtracting $v$ over the common denominator $2v$ leaves $1 - v^2$ on top.
   Section 9.4.2, equations (2) to (4). Lesson: put y = vx

4. Separate $v$ and $x$ (for $v^2 \ne 1$⁠)

   $$\begin{aligned} \frac{2v}{1 - v^2}\,dv &= \frac{dx}{x}, \quad\text{or} \\ \frac{2v}{v^2 - 1}\,dv &= -\frac{dx}{x}. \end{aligned}$$
   Dividing by $\frac{1 - v^2}{2v}$ needs $v^2 \ne 1$⁠; the case $v^2 = 1$ is handled in Step 8. Changing the sign of top and bottom on the left (and moving the minus to the right) makes the top $2v$ the derivative of the bottom $v^2 - 1$⁠.
   Separating, as in Section 9.4.1. Basics: dy = y′ dx

5. Integrate, writing the constant as $\log|C_1|$

   $$\begin{aligned} \int\frac{2v}{v^2 - 1}\,dv &= -\int\frac{1}{x}\,dx \\ \log|v^2 - 1| &= -\log|x| + \log|C_1|, \end{aligned}$$

   where $C_1 \ne 0$ is an arbitrary constant.

   The left integral is $\int\frac{f'(v)}{f(v)}\,dv = \log|f(v)|$ with $f(v) = v^2 - 1$⁠. Any real constant can be written $\log|C_1|$⁠, and this form lets the logs combine in the next step. Still one arbitrary constant.
   Section 7.3.1 (substitution) and Section 7.2, formula (xii). Basics: every method ends with an integral

6. Remove the logs

   $$\begin{aligned} \log|v^2 - 1| + \log|x| &= \log|C_1| \\ \log|(v^2 - 1)x| &= \log|C_1| \\ |(v^2 - 1)x| &= |C_1| \\ (v^2 - 1)x &= \pm C_1. \end{aligned}$$
   $\log a + \log b = \log ab$ for positive $a, b$⁠, and $\log$ is one-one, so equal logs mean equal arguments. Two numbers with equal absolute values are equal or opposite, hence the $\pm$⁠.
   Laws of logarithms; $\log$ is one-one (Chapter 5, Section 5.4). Basics: one constant for order 1

7. Put $v = \frac{y}{x}$ back and rename the constant

   $$\begin{aligned} \left(\frac{y^2}{x^2} - 1\right)x &= \pm C_1 \\ \frac{y^2 - x^2}{x} &= \pm C_1 \\ y^2 - x^2 &= \pm C_1x \\ x^2 - y^2 &= cx, \quad\text{where } c = \mp C_1. \end{aligned}$$
   Multiply by $x \ne 0$⁠, then multiply both sides by $-1$⁠. The new letter $c = \mp C_1$ is again one arbitrary nonzero constant; calling it $c$ gives exactly the family in the question.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

8. The case $v^2 = 1$⁠: the lines $y = \pm x$

   Step 4 set aside $v = \pm 1$⁠, the lines $y = x$ and $y = -x$ (without the origin). On $y = x$ the slope rule gives $\frac{x^2 + x^2}{2x\cdot x} = 1$⁠, and the line has slope 1; on $y = -x$ it gives $\frac{2x^2}{-2x^2} = -1$⁠, the slope of that line. So both lines belong to the family, and they are $x^2 - y^2 = 0\cdot x$⁠, the case $c = 0$⁠.

   Dividing by an expression can lose solutions on which it is zero, so they must be checked separately. Here nothing is lost from the final answer: allowing $c = 0$ in $x^2 - y^2 = cx$ brings them back. The book does not mention this case.
   Every solution, including those set aside when dividing. Lesson: general and particular solutions

9. Check by differentiating back

   Differentiate $x^2 - y^2 = cx$ with respect to $x$⁠: $2x - 2y\,y' = c$⁠. Put $c = \frac{x^2 - y^2}{x}$ (from the curve itself):

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

   which is (1), for every $c$⁠.

   Eliminating $c$ between the curve and its derivative recovers the slope rule, so every curve of the family has the given slope. The picture shows each curve following the little slope marks.
   Implicit differentiation (Chapter 5, Section 5.3.2). Basics: a rule for slopes

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