Exercise 9.5 · Q5: Linear: $\cos^2x\frac{dy}{dx} + y = \tan x$

Basics you need

Solution

1. Divide by $\cos^2x$ to reach the standard form

   $$\frac{dy}{dx} + \sec^2x\cdot y = \sec^2x\tan x.$$
   The standard form needs $\frac{dy}{dx}$ with coefficient 1. On $0 \le x < \frac{\pi}{2}$⁠, $\cos x > 0$⁠, so $\cos^2x \ne 0$ and dividing is allowed; $\frac{1}{\cos^2x} = \sec^2x$ and $\frac{\tan x}{\cos^2x} = \sec^2x\tan x$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

2. Read off $P$ and $Q$

   $P = \sec^2x$ and $Q = \sec^2x\tan x$⁠.

   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. The integrating factor: $e^{\tan x}$

   $$\text{I.F.} = e^{\int\sec^2x\,dx} = e^{\tan x}.$$
   $\int\sec^2x\,dx = \tan x$ (any one antiderivative will do). No modulus appears here, because the antiderivative is not a logarithm.
   Section 9.4.3, step (ii); Section 7.2, formula (iv). Basics: exponentials

4. Multiply and integrate, with one constant

   $$\begin{aligned} \frac{d}{dx}\left(ye^{\tan x}\right) &= \sec^2x\tan x\,e^{\tan x}, \\ ye^{\tan x} &= \int\sec^2x\tan x\,e^{\tan x}\,dx \\ &\qquad + C.\quad\ldots(1) \end{aligned}$$
   By the product rule, $\frac{d}{dx}\left(ye^{\tan x}\right) = e^{\tan x}\frac{dy}{dx} + y\,e^{\tan x}\sec^2x$⁠, which is the left side times the I.F. Integrating gives $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠; this $C$ is the one constant.
   Section 9.4.3, step (iii); product rule and chain rule (Chapter 5, Sections 5.3, 5.3.1). Basics: (y · g)′ = g(y′ + Py)

5. The integral: $t = \tan x$⁠, then by parts

   Put $\tan x = t$⁠, so $\sec^2x\,dx = dt$⁠:

   $$\int\sec^2x\tan x\,e^{\tan x}\,dx = \int t\,e^t\,dt.$$

   By parts, with $t$ as the first function and $e^t$ as the second:

   $$\begin{aligned} \int t\,e^t\,dt &= t\,e^t - \int 1\cdot e^t\,dt \\ &= t\,e^t - e^t = e^t(t - 1) \\ &= e^{\tan x}(\tan x - 1). \end{aligned}$$
   $t = \tan x$ is a legitimate substitution on the interval ($\tan x$ is differentiable there). In the by-parts step, $t$ is the algebraic factor, so it is the first function: its derivative is 1 and the new integral $\int e^t\,dt$ is easy. Check: $\frac{d}{dt}\left[e^t(t - 1)\right] = e^t(t - 1) + e^t = t\,e^t$⁠.
   Section 7.3.1 (substitution); Section 7.6 (by parts). Basics: every method ends with an integral

6. Solve for $y$

   Putting the integral into (1):

   $$\begin{aligned} ye^{\tan x} &= e^{\tan x}(\tan x - 1) + C \\ y &= (\tan x - 1) + Ce^{-\tan x}. \end{aligned}$$
   Divide by $e^{\tan x}$⁠, which is never zero; $\frac{C}{e^{\tan x}} = Ce^{-\tan x}$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

7. Check by substituting back

   From $y = \tan x - 1 + Ce^{-\tan x}$⁠: $\frac{dy}{dx} = \sec^2x - C\sec^2x\,e^{-\tan x}$⁠. Multiply by $\cos^2x$ ($\cos^2x\sec^2x = 1$⁠) and add $y$⁠:

   $$\begin{aligned} &\cos^2x\frac{dy}{dx} + y \\ &= 1 - Ce^{-\tan x} \\ &\quad + \tan x - 1 + Ce^{-\tan x} \\ &= \tan x, \end{aligned}$$

   for every $C$⁠.

   Everything except $\tan x$ cancels, so the family satisfies the given equation itself, not only the divided form.
   A solution makes L.H.S. = R.H.S. (Section 9.3); chain rule (Chapter 5, Section 5.3.1). Lesson: general and particular solutions

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