Exercise 9.3 · Q6: General solution of $\frac{dy}{dx} = (1 + x^2)(1 + y^2)$

Basics you need

Solution

1. Recognise the form: variables separable

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

   with $g(x) = 1 + x^2$ 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, allowed only where $h(y) \ne 0$⁠. Here that is every real $y$⁠, so no solution is lost.
   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 $dx$⁠:

   $$\frac{dy}{1 + y^2} = (1 + x^2)\,dx.$$
   Now $y$ appears only with $dy$ and $x$ only with $dx$⁠, so each side can be integrated on its own.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Integrate both sides, with one constant

   $$\int\frac{dy}{1 + y^2} = \int(1 + x^2)\,dx$$ $$\tan^{-1}y = x + \frac{x^3}{3} + C.$$

   Check each integral by differentiating: $\frac{d}{dy}\tan^{-1}y = \frac{1}{1 + y^2}$ and $\frac{d}{dx}\left(x + \frac{x^3}{3}\right) = 1 + x^2$⁠.

   The left side is the standard $\tan^{-1}$ integral; on the right, the power rule gives $\int 1\,dx = x$ and $\int x^2\,dx = \frac{x^3}{3}$⁠. The constants of integration combine into one $C$⁠.
   Section 7.2, formulas (i) and (x). Basics: every method ends with an integral

4. Check: differentiate the answer back

   Differentiate $\tan^{-1}y = x + \frac{x^3}{3} + C$ with respect to $x$⁠, with $y$ a function of $x$⁠:

   $$\begin{aligned} \frac{1}{1 + y^2}\cdot\frac{dy}{dx} &= 1 + x^2 \\ \Rightarrow\ \frac{dy}{dx} &= (1 + x^2)(1 + y^2). \end{aligned}$$
   We get back the given equation for every value of $C$⁠, so every member of the family is a solution (Section 9.3).
   Chain rule; $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ (Chapter 5). Basics: y′ is the slope

5. $y$ explicitly, and where it holds

   Take tan of both sides:

   $$y = \tan\left(x + \frac{x^3}{3} + C\right),$$

   for the $x$ with $-\frac{\pi}{2} < x + \frac{x^3}{3} + C < \frac{\pi}{2}$⁠.

   $\tan^{-1}y$ is always strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$⁠, so the relation can only hold where the right side is too. Since $x + \frac{x^3}{3}$ increases with $x$⁠, that is one interval of $x$ for each $C$⁠, and $y$ runs off to $\pm\infty$ at its ends. The book leaves the answer in the $\tan^{-1}$ form, which is the general solution.
   Principal value branch of $\tan^{-1}$ (Chapter 2). Lesson: general and particular solutions

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