Example 22: Linear in $x$⁠: $(\tan^{-1}y - x)\,dy = (1 + y^2)\,dx$

Basics you need

Solution

1. Write it with $\frac{dx}{dy}$

   $$\begin{aligned} \frac{dx}{dy} &= \frac{\tan^{-1}y - x}{1 + y^2} \\ \frac{dx}{dy} + \frac{x}{1 + y^2} &= \frac{\tan^{-1}y}{1 + y^2}.\quad\ldots(1) \end{aligned}$$
   Divide by $(1 + y^2)\,dy$⁠; $1 + y^2 \ge 1$ is never 0. Then move the $x$ term to the left. In this form $x$ and $\frac{dx}{dy}$ appear to the first power only, so the equation is linear in $x$⁠.
   Section 9.4.3: the form $\frac{dx}{dy} + P_1x = Q_1$⁠. Lesson: linear equations

2. Read off $P_1$ and $Q_1$

   $P_1 = \frac{1}{1 + y^2}$ and $Q_1 = \frac{\tan^{-1}y}{1 + y^2}$⁠.

   Both are functions of $y$ only, defined for every real $y$⁠, as the form requires.
   Section 9.4.3: $P_1$⁠, $Q_1$ constants or functions of $y$ only. Lesson: linear equations

3. The integrating factor: $e^{\tan^{-1}y}$

   $$\text{I.F.} = e^{\int\frac{1}{1 + y^2}\,dy} = e^{\tan^{-1}y}.$$
   $\int\frac{dy}{1 + y^2} = \tan^{-1}y$⁠; any one antiderivative will do, since an added constant would only multiply the I.F. by a nonzero constant.
   Section 9.4.3: I.F. $= e^{\int P_1\,dy}$⁠; Section 7.2, formula (x). Basics: exponentials

4. Multiply by the I.F.: the left side is one derivative

   $$\frac{d}{dy}\left(x\,e^{\tan^{-1}y}\right) = \frac{\tan^{-1}y}{1 + y^2}\,e^{\tan^{-1}y}.$$
   By the product and chain rules, $\frac{d}{dy}\left(x\,e^{\tan^{-1}y}\right)$ $= e^{\tan^{-1}y}\frac{dx}{dy} + x\cdot\frac{e^{\tan^{-1}y}}{1 + y^2}$⁠, which is the left side of (1) multiplied by $e^{\tan^{-1}y}$⁠.
   Product and chain rules (Chapter 5, Sections 5.3 and 5.3.1). Basics: (y · g)′ = g(y′ + Py)

5. Integrate with respect to $y$⁠, with one constant

   $$\begin{aligned} x\,e^{\tan^{-1}y} &= \int\frac{\tan^{-1}y}{1 + y^2}\,e^{\tan^{-1}y}\,dy \\ &\quad + C.\quad\ldots(2) \end{aligned}$$
   This is $x\,(\text{I.F.}) = \int(Q_1\times\text{I.F.})\,dy + C$⁠. The $C$ here is the one arbitrary constant; the integral below gets none of its own.
   Section 9.4.3: the solution of $\frac{dx}{dy} + P_1x = Q_1$⁠. Basics: one constant for order 1

6. The integral: substitute $t = \tan^{-1}y$⁠, then by parts

   Let $I = \int\frac{\tan^{-1}y}{1 + y^2}\,e^{\tan^{-1}y}\,dy$⁠. Put $\tan^{-1}y = t$⁠; then $\frac{1}{1 + y^2}\,dy = dt$⁠, and

   $$I = \int t\,e^t\,dt.$$

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

   $$\begin{aligned} I &= t\,e^t - \int 1\cdot e^t\,dt \\ &= t\,e^t - e^t = e^t(t - 1) \\ &= e^{\tan^{-1}y}(\tan^{-1}y - 1). \end{aligned}$$
   $\tan^{-1}y$ and its derivative $\frac{1}{1 + y^2}$ both appear in the integrand, which is what makes the substitution work; $t = \tan^{-1}y$ takes every value in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and is one-one, so the substitution is valid for all real $y$⁠. In $\int te^t\,dt$ the algebraic factor $t$ is the first function (ILATE), because differentiating it gives 1.
   Section 7.3.1, substitution; Section 7.6, integration by parts. Basics: every method ends with an integral

7. Solve for $x$

   Putting $I$ into (2):

   $$\begin{aligned} x\,e^{\tan^{-1}y} &= e^{\tan^{-1}y}(\tan^{-1}y - 1) + C \\ x &= (\tan^{-1}y - 1) + Ce^{-\tan^{-1}y}. \end{aligned}$$
   Divide by $e^{\tan^{-1}y}$⁠, which is never 0; $\frac{C}{e^{\tan^{-1}y}} = Ce^{-\tan^{-1}y}$⁠. The answer holds for every real $y$⁠.
   Section 9.4.3: the general solution. Lesson: Examples 19 to 22

8. Check by substituting back

   Write $t = \tan^{-1}y$⁠, so $x = t - 1 + Ce^{-t}$ and $\frac{dt}{dy} = \frac{1}{1 + y^2}$⁠. Then

   $$\frac{dx}{dy} = \left(1 - Ce^{-t}\right)\frac{1}{1 + y^2},$$ $$\begin{aligned} &\frac{dx}{dy} + \frac{x}{1 + y^2} \\ &= \frac{1 - Ce^{-t} + t - 1 + Ce^{-t}}{1 + y^2} \\ &= \frac{t}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}, \end{aligned}$$

   which is (1), for every $C$⁠.

   The $C$ terms cancel, as they must: $Ce^{-\tan^{-1}y}$ alone solves $\frac{dx}{dy} + \frac{x}{1 + y^2} = 0$⁠, and $\tan^{-1}y - 1$ supplies the right side.
   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

Every fact used, and where it comes from