Exercise 9.5 · Q13: $\frac{dy}{dx} + 2y\tan x = \sin x$⁠, $y(\frac{\pi}{3}) = 0$

Basics you need

Solution

1. Recognise the linear form: $P = 2\tan x$⁠, $Q = \sin x$

   The equation already has the form $\frac{dy}{dx} + Py = Q$ with

   $$P = 2\tan x, \qquad Q = \sin x.$$
   The equation has order 1 and degree 1: $y$ and $\frac{dy}{dx}$ appear to the first power, and $P$⁠, $Q$ depend on $x$ only. $\tan x$ needs $\cos x \ne 0$⁠. The given point $x = \frac{\pi}{3}$ lies in the interval $-\frac{\pi}{2} < x < \frac{\pi}{2}$⁠, and we work there.
   Section 9.4.3, step (i). Lesson: linear equations

2. The integrating factor: $\sec^2x$

   $$\begin{aligned} \text{I.F.} &= e^{\int 2\tan x\,dx} = e^{2\log|\sec x|} \\ &= e^{\log\sec^2x} = \sec^2x. \end{aligned}$$
   $2\log|\sec x| = \log|\sec x|^2 = \log\sec^2x$⁠, and $e^{\log t} = t$⁠. Because of the square there is no sign question at all: $|\sec x|^2 = \sec^2x$ on every interval.
   Section 9.4.3, step (ii); Section 7.3.1, $\int\tan x\,dx$⁠. Basics: exponentials

3. Multiply by $\sec^2x$⁠: the left side is one derivative

   $$\begin{aligned} &\sec^2x\frac{dy}{dx} + 2\sec^2x\tan x\cdot y \\ &\quad = \sin x\sec^2x, \\ &\frac{d}{dx}\left(y\sec^2x\right) = \sec x\tan x. \end{aligned}$$
   By the chain rule, $\frac{d}{dx}\sec^2x = 2\sec x\cdot\sec x\tan x = 2\sec^2x\tan x$⁠, so the product rule gives exactly the left side. On the right, $\sin x\sec^2x = \frac{\sin x}{\cos x}\cdot\frac{1}{\cos x} = \tan x\sec x$⁠.
   Product and chain rules (Chapter 5, Sections 5.3 and 5.3.1). Basics: (y · g)′ = g(y′ + Py)

4. Integrate, with one constant

   $$\begin{aligned} y\sec^2x &= \int\sin x\sec^2x\,dx + C \\ &= \int\sec x\tan x\,dx + C \\ &= \sec x + C. \end{aligned}$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠. The $C$ is the one arbitrary constant of the general solution.
   Section 9.4.3, step (iii); Section 7.2, formula (vi). Basics: one constant for order 1

5. Solve for $y$⁠: the general solution

   $$y = \cos x + C\cos^2x.\quad\ldots(1)$$
   Divide by $\sec^2x$⁠, that is, multiply by $\cos^2x$⁠: $\sec x\cdot\cos^2x = \cos x$⁠. This is the general solution, a family of curves with one constant.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

6. Use the condition: $y = 0$ when $x = \frac{\pi}{3}$

   Putting $x = \frac{\pi}{3}$⁠, $y = 0$ in (1):

   $$\begin{aligned} 0 &= \cos\frac{\pi}{3} + C\cos^2\frac{\pi}{3} = \frac{1}{2} + \frac{C}{4} \\ C &= -2. \end{aligned}$$
   $\cos\frac{\pi}{3} = \frac{1}{2}$⁠, so $\cos^2\frac{\pi}{3} = \frac{1}{4}$⁠. Multiplying $0 = \frac{1}{2} + \frac{C}{4}$ by 4 gives $0 = 2 + C$⁠.
   Section 9.3: a particular solution. Lesson: general and particular solutions

7. The particular solution

   $$y = \cos x - 2\cos^2x.$$
   Put $C = -2$ in (1). The formula is defined for every $x$⁠, but the equation is not defined where $\cos x = 0$⁠; the solution through $\left(\frac{\pi}{3}, 0\right)$ lives on $-\frac{\pi}{2} < x < \frac{\pi}{2}$⁠.
   Section 9.3: a particular solution. Lesson: Examples 14 to 18

8. Check by substituting back, and see the curve

   From $y = \cos x - 2\cos^2x$⁠: $\frac{dy}{dx} = -\sin x + 4\sin x\cos x$⁠. And

   $$\begin{aligned} 2y\tan x &= 2(\cos x - 2\cos^2x)\frac{\sin x}{\cos x} \\ &= 2\sin x - 4\sin x\cos x. \end{aligned}$$

   Adding, $\frac{dy}{dx} + 2y\tan x = \sin x$⁠, the given equation. At $x = \frac{\pi}{3}$⁠: $y = \frac{1}{2} - 2\cdot\frac{1}{4} = 0$⁠.

   The $4\sin x\cos x$ terms cancel and $-\sin x + 2\sin x = \sin x$⁠, so the formula satisfies the equation wherever $\cos x \ne 0$⁠, and it passes through the given point.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Basics: a rule for slopes

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