Exercise 9.3 · Q1: General solution of $\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}$

Basics you need

Solution

1. Recognise the form: variables separable

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

   with $g(x) = \frac{1 - \cos x}{1 + \cos x}$ and $h(y) = 1$⁠. The equation only makes sense where $1 + \cos x \ne 0$⁠, that is $x \ne \pm\pi, \pm 3\pi, \ldots$

   An equation $\frac{dy}{dx} = g(x)$ is the simplest separable case: $h(y) = 1$ is never 0, so there is nothing to divide by and no solution can be lost. The only restriction comes from the equation itself, whose denominator $1 + \cos x$ must not vanish.
   Section 9.4.1: $\frac{dy}{dx} = h(y)\,g(x)$⁠, with $h(y) \ne 0$⁠. Lesson: separate, then integrate

2. Separate: $dy$ in terms of $dx$

   Multiply both sides by $dx$⁠:

   $$dy = \frac{1 - \cos x}{1 + \cos x}\,dx.$$
   Now the left side holds only $y$ (with $dy$⁠) and the right side only $x$ (with $dx$⁠). Integrating the left side will give $y$ itself.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Simplify with the half-angle formulas

   $$\begin{aligned} \frac{1 - \cos x}{1 + \cos x} &= \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2} \\ &= \sec^2\frac{x}{2} - 1. \end{aligned}$$
   The fraction as it stands has no standard antiderivative. Writing both $1 \pm \cos x$ through the half angle cancels the 2s and leaves $\tan^2\frac{x}{2}$⁠. Then $\tan^2 = \sec^2 - 1$ turns it into $\sec^2$⁠, whose integral is on the standard list. Where $1 + \cos x \ne 0$ we have $\cos\frac{x}{2} \ne 0$⁠, so $\tan\frac{x}{2}$ and $\sec\frac{x}{2}$ are defined.
   $\cos 2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$⁠; $1 + \tan^2\theta = \sec^2\theta$ (Class XI). Basics: every method ends with an integral

4. Integrate both sides, with one constant

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

   Check each integral by differentiating: $\frac{d}{dx}\left(2\tan\frac{x}{2}\right) = 2\sec^2\frac{x}{2}\cdot\frac{1}{2} = \sec^2\frac{x}{2}$ and $\frac{d}{dx}(x) = 1$⁠.

   From $\int\sec^2u\,du = \tan u$ with $u = \frac{x}{2}$⁠, $du = \frac{1}{2}dx$⁠, so $dx = 2\,du$ and $\int\sec^2\frac{x}{2}\,dx = 2\tan\frac{x}{2}$⁠. The constants of integration from both sides combine into one $C$⁠, which may be any real number: one constant for a first order equation.
   Section 7.2, formula (iv), with the substitution $u = \frac{x}{2}$ (Section 7.3.1). Basics: order n gives n constants

5. Check: substitute back

   For $y = 2\tan\frac{x}{2} - x + C$⁠,

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

   reading Step 3 backwards. So L.H.S. = R.H.S. for every $C$ and every $x$ with $\cos x \ne -1$⁠.

   Substituting back is the definition of a solution (Section 9.3). The solution lives on each interval between consecutive odd multiples of $\pi$⁠, for example $-\pi < x < \pi$⁠, where $\tan\frac{x}{2}$ is defined; as $x \to \pi$ the curve goes off to infinity, just as the slope does.
   A solution makes L.H.S. = R.H.S. Lesson: general and particular solutions

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