Example 5: General solution of $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$

Basics you need

Solution

1. Recognise the form: variables separable

   $$\frac{dy}{dx} = \frac{1}{1 + x^2}\cdot(1 + y^2) = g(x)\,h(y),$$

   with $g(x) = \frac{1}{1 + x^2}$ and $h(y) = 1 + y^2$⁠. For every real $y$⁠, $1 + y^2 \ge 1$⁠, so $h(y) \ne 0$⁠.

   Dividing by $h(y)$ is the separating step, and it is only allowed where $h(y) \ne 0$⁠. Here that is every real $y$⁠, so no solution is lost by dividing. (When $h$ can be 0, as in Example 6, the zeros need a separate look.)
   Section 9.4.1: $\frac{dy}{dx} = h(y)\,g(x)$⁠, with $h(y) \ne 0$⁠. Lesson: separate, then integrate

2. Separate the variables

   Divide both sides by $1 + y^2$ and multiply by $dx$⁠:

   $$\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}.$$
   Now each side holds one variable only: $y$ with $dy$ on the left, $x$ with $dx$ on the right. Both denominators are at least 1, so nothing here is undefined.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Integrate both sides, with one constant

   $$\int\frac{dy}{1 + y^2} = \int\frac{dx}{1 + x^2}$$ $$\tan^{-1}y = \tan^{-1}x + C.$$
   Both integrals are the standard one, $\int\frac{dt}{1 + t^2} = \tan^{-1}t$⁠, once with $t = y$ and once with $t = x$⁠. The two constants of integration combine into one $C$⁠, which may be any real number. This is the general solution.
   Section 7.2, formula (x): $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$⁠. Basics: every method ends with an integral

4. Check: differentiate the answer back

   Differentiate $\tan^{-1}y = \tan^{-1}x + C$ with respect to $x$⁠, with $y$ a function of $x$ (chain rule on the left):

   $$\begin{aligned} \frac{1}{1 + y^2}\cdot\frac{dy}{dx} &= \frac{1}{1 + x^2} \\ \Rightarrow\ \frac{dy}{dx} &= \frac{1 + y^2}{1 + x^2}. \end{aligned}$$
   We get back exactly the given equation, whatever the value of $C$⁠. So every member of the family is a solution.
   Chain rule and the derivative of $\tan^{-1}$ (Chapter 5). Basics: y′ is the slope

5. Which constants give curves, and the curves without $\tan^{-1}$

   Both $\tan^{-1}y$ and $\tan^{-1}x$ lie in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, so $C = \tan^{-1}y - \tan^{-1}x$ always lies in $(-\pi, \pi)$⁠. Only these values of $C$ give curves.

   For $C \ne \pm\frac{\pi}{2}$⁠, put $k = \tan C$ and take tan of both sides of $\tan^{-1}y = \tan^{-1}x + C$⁠. The formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$ gives

   $$y = \frac{x + k}{1 - kx},$$

   for the $x$ with $\tan^{-1}x + C$ inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. For example $C = 0$ gives the line $y = x$⁠, and $C = \frac{\pi}{4}$ ($k = 1$⁠) gives $y = \frac{1 + x}{1 - x}$ for $x < 1$⁠.

   The $\tan^{-1}$ form is the general solution the book asks for; this step only shows what its curves are. The restriction on $x$ is needed because $\tan^{-1}y$ can never reach $\pm\frac{\pi}{2}$⁠: for $C = \frac{\pi}{4}$⁠, the solution through $(0, 1)$ blows up as $x \to 1$⁠.
   $\tan(A + B)$ formula (Class XI); range of $\tan^{-1}$ (Chapter 2). Lesson: general and particular solutions

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