Exercise 9.5 · Q4: Linear: $\frac{dy}{dx} + (\sec x)y = \tan x$

Basics you need

Solution

1. Recognise the linear form: $P = \sec x$⁠, $Q = \tan x$

   The equation already has the form $\frac{dy}{dx} + Py = Q$ with $P = \sec x$ and $Q = \tan x$⁠.

   $y$ and $\frac{dy}{dx}$ appear to the first power, and $P$⁠, $Q$ depend on $x$ only. Both are defined on $0 \le x < \frac{\pi}{2}$⁠, where $\cos x > 0$⁠; the interval stops before $\frac{\pi}{2}$⁠, where $\sec x$ and $\tan x$ are undefined.
   Section 9.4.3, step (i). Lesson: linear equations

2. The integrating factor: $\sec x + \tan x$

   $$\begin{aligned} \text{I.F.} &= e^{\int\sec x\,dx} \\ &= e^{\log|\sec x + \tan x|} \\ &= \sec x + \tan x. \end{aligned}$$
   On $0 \le x < \frac{\pi}{2}$⁠, $\sec x \ge 1$ and $\tan x \ge 0$⁠, so $\sec x + \tan x \ge 1 > 0$ and the modulus can be dropped; then $e^{\log t} = t$⁠.
   Section 9.4.3, step (ii); Section 7.3.1. Basics: exponentials

3. Multiply by $\sec x + \tan x$⁠: the left side is one derivative

   $$\begin{aligned} &(\sec x + \tan x)\frac{dy}{dx} \\ &\quad + \sec x(\sec x + \tan x)\,y \\ &= \tan x(\sec x + \tan x), \end{aligned}$$

   that is,

   $$\begin{aligned} &\frac{d}{dx}\big[y(\sec x + \tan x)\big] \\ &= \sec x\tan x + \tan^2x. \end{aligned}$$
   The derivative of $\sec x + \tan x$ is $\sec x\tan x + \sec^2x$ $= \sec x(\tan x + \sec x)$⁠, so by the product rule $\frac{d}{dx}\big[y(\sec x + \tan x)\big]$ is exactly the left side.
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

4. Integrate, with one constant

   $$\begin{aligned} &y(\sec x + \tan x) \\ &= \int(\sec x\tan x + \tan^2x)\,dx \\ &\qquad + C \\ &= \int(\sec x\tan x + \sec^2x - 1)\,dx \\ &\qquad + C \\ &= \sec x + \tan x - x + C. \end{aligned}$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠. The identity $\tan^2x = \sec^2x - 1$ replaces $\tan^2x$⁠, which has no formula of its own in the list, by two standard integrands. Then term by term: $\int\sec x\tan x\,dx = \sec x$⁠, $\int\sec^2x\,dx = \tan x$⁠, $\int 1\,dx = x$⁠. Check: $\frac{d}{dx}(\sec x + \tan x - x)$ $= \sec x\tan x + \sec^2x - 1$⁠.
   Section 9.4.3, step (iii); Section 7.2, formulas (vi) and (iv). Basics: every method ends with an integral

5. The general solution, and $y$ on its own

   $$\begin{aligned} &y(\sec x + \tan x) \\ &= \sec x + \tan x - x + C. \end{aligned}$$

   Dividing by $\sec x + \tan x$ (positive), $y = 1 + \dfrac{C - x}{\sec x + \tan x}$⁠.

   The book leaves the answer in the first form, $y\,(\text{I.F.}) = \ldots$⁠; the second form is the same equation solved for $y$⁠, and it is the one used in the check.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

6. Check by substituting back

   Write $g = \sec x + \tan x$⁠, so $g' = \sec x\cdot g$ and $\frac{1}{g} = \sec x - \tan x$ (because $\sec^2x - \tan^2x = 1$⁠). From $y = 1 + (C - x)\frac{1}{g}$⁠:

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

   Adding,

   $$\begin{aligned} &\frac{dy}{dx} + (\sec x)\,y \\ &= \sec x - \frac{1}{g} \\ &= \sec x - (\sec x - \tan x) \\ &= \tan x, \end{aligned}$$

   for every $C$⁠.

   The $C$ terms cancel, and what is left is $\sec x - \frac{1}{\sec x + \tan x}$⁠, which equals $\tan x$ because $(\sec x + \tan x)(\sec x - \tan x)$ $= \sec^2x - \tan^2x = 1$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3); quotient rule (Chapter 5). Lesson: general and particular solutions

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