Miscellaneous · Q4: General solution of $\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$

Basics you need

Solution

1. Where the equation makes sense; split the root

   The denominator $1 - x^2$ must be positive (it is under a root and below a fraction line), so $-1 < x < 1$⁠. Then the quotient under the root is $\ge 0$ only if $1 - y^2 \ge 0$⁠, that is $-1 \le y \le 1$⁠. With $1 - y^2 \ge 0$ and $1 - x^2 > 0$⁠,

   $$\frac{dy}{dx} = -\sqrt{\frac{1 - y^2}{1 - x^2}} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}.$$
   The form $\frac{dy}{dx} = h(y)\,g(x)$ needs the root split into a $y$-part and an $x$-part. That split is valid exactly because the top is not negative and the bottom is positive.
   $\sqrt{p/q} = \sqrt p/\sqrt q$ for $p \ge 0$⁠, $q > 0$⁠. Basics: dy = y′ dx

2. Separate the variables

   For $-1 < y < 1$⁠, $\sqrt{1 - y^2} \ne 0$⁠, and we may divide by it:

   $$\frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}}.$$
   Now the left side has only $y$ and the right side only $x$⁠. This is the book's form (3) of Section 9.4.1 with $h(y) = \sqrt{1 - y^2}$ and $g(x) = -\frac{1}{\sqrt{1 - x^2}}$⁠. The two values $y = \pm 1$⁠, where we cannot divide, are looked at in Step 5.
   Section 9.4.1: separating the variables. Lesson: variables separable

3. Integrate, with one constant

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

   where $C$ is an arbitrary constant.

   Both integrals are the same standard form. Each side would bring its own constant, but their difference is one arbitrary number, so one $C$ on the right is enough.
   Section 7.2, formula (viii); Section 9.4.1, equation (4). Basics: every method ends with an integral

4. The general solution, and where it lives

   $$\sin^{-1}x + \sin^{-1}y = C,$$

   for $-1 < x < 1$ and $-1 < y < 1$⁠. Since each inverse sine lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ here, only constants with $-\pi < C < \pi$ give curves. Solved for $y$⁠: $y = \sin(C - \sin^{-1}x)$⁠, for the $x$ with $C - \sin^{-1}x$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠.

   Moving $\sin^{-1}x$ to the left gives the symmetric form of the answer key. The range of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Chapter 2), which is why $C$ is limited and why the explicit form holds only on part of $(-1, 1)$⁠.
   Principal value branch of $\sin^{-1}$ (Chapter 2, Section 2.2). Lesson: Examples 4 to 9

5. The constant solutions $y = 1$ and $y = -1$

   In Step 2 we divided by $\sqrt{1 - y^2}$⁠, which is 0 when $y = \pm 1$⁠. The constant functions $y = 1$ and $y = -1$ have $\frac{dy}{dx} = 0$ and make the root $\sqrt{\frac{0}{1 - x^2}} = 0$⁠, so they also satisfy the equation on $-1 < x < 1$⁠. No value of $C$ gives them: $y = 1$ would need $\sin^{-1}x = C - \frac{\pi}{2}$ for every $x$⁠.

   Dividing by an expression can lose the solutions on which it vanishes, so we test them directly. These two lines lie outside the family; the question asks for the general solution, which is the family of Step 4. The answer key gives the family only.
   Section 9.4.1: separating needs $h(y) \ne 0$⁠, so $y = \pm 1$ are tested directly; chapter Summary: one arbitrary constant in the general solution. Lesson: general and particular solutions

6. Check by substituting back

   Differentiate $\sin^{-1}x + \sin^{-1}y = C$ with respect to $x$⁠:

   $$\begin{aligned} \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} &= 0, \\ \frac{dy}{dx} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} &= -\sqrt{\frac{1 - y^2}{1 - x^2}}, \end{aligned}$$

   which is the given equation, for every $C$⁠.

   Getting the equation back proves the answer. The minus sign comes out right because both roots are positive; on the curve, $\sqrt{1 - y^2} = \cos(\sin^{-1}y) \ge 0$ since $\sin^{-1}y$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠.
   $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$ (Chapter 5, Section 5.3.3). Basics: y′ and y″

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