Exercise 9.3 · Q7: General solution of $y\log y\,dx - x\,dy = 0$

Basics you need

Solution

1. Divide by $xy\log y$⁠: the variables separate

   The equation contains $\log y$⁠, so $y > 0$⁠. Move the $dy$ term across: $y\log y\,dx = x\,dy$⁠. For $x \ne 0$ and $y \ne 1$ (so $\log y \ne 0$⁠), divide by $xy\log y$⁠:

   $$\frac{dx}{x} = \frac{dy}{y\log y}.$$
   Dividing needs $x \ne 0$⁠, $y \ne 0$ and $\log y \ne 0$⁠; the first two are clear and $\log y = 0$ only at $y = 1$⁠. The constant function $y = 1$ does satisfy the equation ($1\cdot\log 1 = 0$ and $dy = 0$⁠), so we must remember it; it returns in Step 4.
   Section 9.4.1: separate by dividing by a nonzero factor. Lesson: separate, then integrate

2. Integrate both sides, with one constant $\log|C|$

   On the left put $t = \log y$⁠, so $dt = \frac{1}{y}\,dy$⁠:

   $$\begin{aligned} \int\frac{dy}{y\log y} &= \int\frac{dt}{t} = \log|t| \\ &= \log|\log y|. \end{aligned}$$

   So, with the constant written as $\log|C|$ ($C \ne 0$⁠),

   $$\log|\log y| = \log|x| + \log|C|.$$

   Check each integral by differentiating: $\frac{d}{dy}\log|\log y| = \frac{1}{\log y}\cdot\frac{1}{y}$⁠, and $\frac{d}{dx}\log|x| = \frac{1}{x}$⁠.

   $\frac{1}{y}$ is the derivative of $\log y$⁠, so the left side is "derivative over function". Both sides being logs, the constant is written $\log|C|$⁠; as $C$ runs over the nonzero reals, $\log|C|$ takes every real value, so it is one arbitrary constant.
   Section 7.2, formula (xii), with a substitution (Section 7.3.1). Basics: every method ends with an integral

3. Remove the logs: solve for $y$

   $$\begin{aligned} \log|\log y| &= \log|Cx| \\ \Rightarrow\ |\log y| &= |Cx| \\ \Rightarrow\ \log y &= Cx, \end{aligned}$$

   writing $C$ again for $\pm C$⁠. Taking the exponential,

   $$y = e^{Cx}.$$
   $\log$ is one-one, so equal logs mean equal arguments. Along a solution $\frac{\log y}{x}$ is continuous and never 0 here, so its sign does not change and $\pm$ is one fixed sign, absorbed into the arbitrary $C$⁠.
   $\log|a| + \log|b| = \log|ab|$⁠; $e^{\log y} = y$⁠. Basics: the exponential function

4. Bring back $y = 1$⁠: the case $C = 0$

   Putting $C = 0$ in $y = e^{Cx}$ gives $y = e^0 = 1$⁠, the solution set aside in Step 1. So $y = e^{Cx}$ with $C$ any real number covers it.

   The derivation needed $C \ne 0$ (to write $\log|C|$⁠), but the final formula makes sense for $C = 0$ too, and that value gives exactly the lost solution. So the one formula is the complete general solution, as the textbook states it.
   Section 9.3: general solution with one arbitrary constant. Basics: one arbitrary constant

5. Check: substitute back

   For $y = e^{Cx}$⁠: $\log y = Cx$ and $\frac{dy}{dx} = Ce^{Cx}$⁠, so $dy = Ce^{Cx}\,dx$ and

   $$\begin{aligned} &y\log y\,dx - x\,dy \\ &= e^{Cx}\cdot Cx\,dx - x\cdot Ce^{Cx}\,dx = 0. \end{aligned}$$
   L.H.S. = R.H.S. for every $C$ and every $x$⁠, including $C = 0$⁠. So every member of the family is a solution (Section 9.3), and $y = e^{Cx} > 0$⁠, as $\log y$ needs.
   A solution makes L.H.S. = R.H.S. Lesson: general and particular solutions

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