Exercise 9.3 · Q2: General solution of $\frac{dy}{dx} = \sqrt{4 - y^2}$⁠, $-2 < y < 2$

Basics you need

Solution

1. Recognise the form: variables separable

   $$\frac{dy}{dx} = 1\cdot\sqrt{4 - y^2} = g(x)\,h(y),$$

   with $g(x) = 1$ and $h(y) = \sqrt{4 - y^2}$⁠. For $-2 < y < 2$ we have $y^2 < 4$⁠, so $4 - y^2 > 0$ and $h(y) > 0$⁠.

   Separating means dividing by $h(y)$⁠, which is allowed only where $h(y) \ne 0$⁠. The zeros of $h$ are $y = \pm 2$⁠, and the question has excluded them. (The constant functions $y = 2$ and $y = -2$ do satisfy the equation, but they lie outside $-2 < y < 2$⁠.)
   Section 9.4.1: separation needs $h(y) \ne 0$⁠. Lesson: separate, then integrate

2. Separate the variables

   Divide both sides by $\sqrt{4 - y^2}$ (positive) and multiply by $dx$⁠:

   $$\frac{dy}{\sqrt{4 - y^2}} = dx.$$
   Now the left side has only $y$ and the right side only $x$⁠. The square root is of a positive number, so the left side is defined all along.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Integrate both sides, with one constant

   $$\begin{aligned} \int\frac{dy}{\sqrt{2^2 - y^2}} &= \int dx \\ \Rightarrow\ \sin^{-1}\frac{y}{2} &= x + C. \end{aligned}$$

   Check each integral by differentiating: $\frac{d}{dx}(x) = 1$⁠, and by the chain rule

   $$\frac{d}{dy}\sin^{-1}\frac{y}{2} = \frac{1}{\sqrt{1 - \frac{y^2}{4}}}\cdot\frac{1}{2} = \frac{1}{\sqrt{4 - y^2}}.$$
   Writing $4 = 2^2$ shows the standard form $\int\frac{dy}{\sqrt{a^2 - y^2}}$ with $a = 2$⁠, valid for $-2 < y < 2$⁠, exactly our range. The two constants of integration combine into one $C$⁠.
   Section 7.4, formula (5). Basics: every method ends with an integral

4. Solve for $y$⁠, and say where it holds

   Take sine of both sides of $\sin^{-1}\frac{y}{2} = x + C$⁠:

   $$\frac{y}{2} = \sin(x + C) \;\Rightarrow\; y = 2\sin(x + C).$$

   Since $\sin^{-1}\frac{y}{2}$ lies strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ (as $-1 < \frac{y}{2} < 1$⁠), the relation holds for the $x$ with

   $$-\frac{\pi}{2} < x + C < \frac{\pi}{2}.$$
   $\sin(\sin^{-1}t) = t$⁠, so taking sine undoes $\sin^{-1}$⁠. This is the book's form of the general solution. The interval matters: on it $y$ rises from $-2$ to $2$⁠. Beyond it the formula $2\sin(x + C)$ starts to fall, while the equation demands a slope $\sqrt{4 - y^2} \ge 0$⁠.
   Principal value branch of $\sin^{-1}$ (Chapter 2). Lesson: general and particular solutions

5. Check: substitute back, and see the curves

   Let $u = x + C$ with $-\frac{\pi}{2} < u < \frac{\pi}{2}$⁠, so $\cos u > 0$⁠. For $y = 2\sin u$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= 2\cos u, \\ \sqrt{4 - y^2} &= \sqrt{4 - 4\sin^2u} = \sqrt{4\cos^2u} \\ &= 2|\cos u| = 2\cos u. \end{aligned}$$

   Both sides agree, and $-2 < y < 2$ there. So L.H.S. = R.H.S. for every $C$⁠.

   The check needs $|\cos u| = \cos u$⁠, which is exactly why the interval $-\frac{\pi}{2} < x + C < \frac{\pi}{2}$ was needed. The picture shows each curve following the slope marks, which are steepest at $y = 0$ and flatten towards $y = \pm 2$⁠.
   A solution makes L.H.S. = R.H.S. Basics: dy/dx = F(x, y) draws an arrow at every point

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