Exercise 9.5 · Q7: Linear: $x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x$

Basics you need

Solution

1. Where the equation lives, and the standard form

   $$\frac{dy}{dx} + \frac{1}{x\log x}\,y = \frac{2}{x^2}.$$
   $\log x$ needs $x > 0$⁠. To divide by $x\log x$ we also need $\log x \ne 0$⁠, that is $x \ne 1$⁠. So we work on $0 < x < 1$ or on $x > 1$⁠. On the right, $\frac{2}{x}\log x \div x\log x = \frac{2}{x^2}$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

2. Read off $P$ and $Q$

   $P = \frac{1}{x\log x}$ and $Q = \frac{2}{x^2}$⁠.

   Both depend on $x$ only, and $y$ appears to the first power, so the equation is linear.
   Section 9.4.3: $P$ and $Q$ are constants or functions of $x$ only. Lesson: linear equations

3. $\int P\,dx$ by the substitution $t = \log x$

   Put $\log x = t$⁠, so $\frac{1}{x}\,dx = dt$⁠:

   $$\begin{aligned} \int\frac{dx}{x\log x} &= \int\frac{dt}{t} \\ &= \log|t| = \log|\log x|. \end{aligned}$$
   $\frac{dx}{x\log x}$ is $\frac{1}{\log x}$ times $\frac{dx}{x}$⁠, and $\frac{dx}{x}$ is exactly $dt$⁠. The modulus stays because $\log x$ is negative on $0 < x < 1$⁠.
   Section 7.3.1 (substitution); Section 7.2, formula (xii). Basics: every method ends with an integral

4. The integrating factor: $\log x$

   $$\text{I.F.} = e^{\log|\log x|} = |\log x|.$$

   We take $\text{I.F.} = \log x$⁠, as the book does.

   For $x > 1$⁠, $\log x > 0$ and $|\log x| = \log x$⁠. For $0 < x < 1$⁠, $|\log x| = -\log x$⁠, a constant multiple ($-1$⁠) of $\log x$⁠, and a nonzero constant multiple of an integrating factor is again one. So $\log x$ works on both intervals. Indeed $\frac{d}{dx}(\log x) = \frac{1}{x} = P\cdot\log x$⁠.
   Section 9.4.3, step (ii). Basics: exponentials

5. Multiply and integrate, with one constant

   $$\begin{aligned} \log x\,\frac{dy}{dx} + \frac{y}{x} &= \frac{2}{x^2}\log x \\ \frac{d}{dx}(y\log x) &= \frac{2}{x^2}\log x \\ y\log x &= \int\frac{2}{x^2}\log x\,dx + C. \end{aligned}$$
   By the product rule, $\frac{d}{dx}(y\log x) = \log x\,\frac{dy}{dx} + y\cdot\frac{1}{x}$⁠, the left side. Integrating gives $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠, with the one constant $C$⁠.
   Section 9.4.3, step (iii); product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

6. $\int\frac{2}{x^2}\log x\,dx$ by parts

   Take $\log x$ as the first function and $\frac{2}{x^2}$ as the second, so $\int\frac{2}{x^2}\,dx = -\frac{2}{x}$⁠:

   $$\begin{aligned} &\int\frac{2}{x^2}\log x\,dx \\ &= \log x\cdot\left(-\frac{2}{x}\right) \\ &\quad - \int\frac{1}{x}\left(-\frac{2}{x}\right)dx \\ &= -\frac{2}{x}\log x + \int\frac{2}{x^2}\,dx \\ &= -\frac{2}{x}\log x - \frac{2}{x} \\ &= -\frac{2}{x}(1 + \log x). \end{aligned}$$
   $\int 2x^{-2}\,dx = \frac{2x^{-1}}{-1} = -\frac{2}{x}$ by the power rule. Check: the derivative of $-\frac{2}{x}(1 + \log x)$ is $\frac{2}{x^2}(1 + \log x) - \frac{2}{x}\cdot\frac{1}{x}$ $= \frac{2}{x^2}\log x$⁠.
   Section 7.6, integration by parts; Section 7.2, formula (i). Basics: every method ends with an integral

7. The general solution

   $$y\log x = -\frac{2}{x}(1 + \log x) + C.$$
   This is the book's form. Solved for $y$⁠: $y = \dfrac{C - \frac{2}{x}(1 + \log x)}{\log x}$⁠, on $0 < x < 1$ or on $x > 1$⁠. At $x = 1$ the given equation reads $0 + y = 0$⁠, so a solution can only reach $x = 1$ with $y = 0$⁠; in the family that happens only for $C = 2$⁠, where $C - \frac{2}{x}(1 + \log x)$ also vanishes at $x = 1$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

8. Check by substituting back

   Differentiate $y\log x = -\frac{2}{x}(1 + \log x) + C$ on both sides:

   $$\begin{aligned} \log x\,\frac{dy}{dx} + \frac{y}{x} &= \frac{2}{x^2}(1 + \log x) - \frac{2}{x^2} \\ &= \frac{2}{x^2}\log x. \end{aligned}$$

   Multiplying by $x$⁠: $x\log x\,\frac{dy}{dx} + y = \frac{2}{x}\log x$⁠, the given equation, for every $C$⁠.

   Differentiating the solution in the form $y\,(\text{I.F.}) = \ldots$ gives back the multiplied equation, and multiplying by $x$ gives the original one. The constant disappears on differentiation, as it must.
   A solution makes L.H.S. = R.H.S. (Section 9.3); product and quotient rules (Chapter 5). Lesson: general and particular solutions

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