Example 38: $y=\sin^{-1}x$ satisfies $(1-x^2)y''-xy'=0$

Basics you need

Solution

1. First derivative, and clear the fraction

   For $-1 \lt x \lt 1$⁠, by Basic 1,

   $$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}},$$

   so, multiplying both sides by $\sqrt{1-x^2}$⁠,

   $$\sqrt{1-x^2}\,\frac{dy}{dx} = 1.$$
   Multiplying by $\sqrt{1-x^2}$ (which is positive on $(-1,1)$⁠) removes the fraction. Differentiating a product is easier than differentiating a quotient containing a square root, and the right side becomes the constant 1, whose derivative is 0.
   On $(-1, 1)$⁠, $1 - x^2 \gt 0$⁠, so $\sqrt{1-x^2} \gt 0$⁠. Lesson: derivatives of inverse trig functions

2. Differentiate both sides with respect to $x$

   The left side is a product $u\,v$ with $u = \sqrt{1-x^2}$ and $v = \frac{dy}{dx}$⁠. By the product rule (Basic 2):

   $$\sqrt{1-x^2}\,\frac{d^2y}{dx^2} + \frac{dy}{dx}\cdot\frac{d}{dx}\sqrt{1-x^2} = 0.$$
   If two expressions are equal for every $x$ in an interval, their derivatives are equal too. The derivative of $v = \frac{dy}{dx}$ is, by definition, $\frac{d^2y}{dx^2}$ (Basic 4). The right side, 1, has derivative 0.
   Product rule (Section 5.3). Basics: derivative rules

3. Differentiate the square root

   By Basic 3, $\frac{d}{dx}\sqrt{1-x^2} = \frac{-x}{\sqrt{1-x^2}}$⁠. Substituting into Step 2:

   $$\sqrt{1-x^2}\,\frac{d^2y}{dx^2} - \frac{x}{\sqrt{1-x^2}}\,\frac{dy}{dx} = 0.$$
   This is the chain rule: the derivative of the outer square root, $\frac{1}{2\sqrt{1-x^2}}$⁠, times the derivative of the inside, $-2x$⁠. The 2s cancel, as in the book's line with $\frac{2x}{2\sqrt{1-x^2}}$⁠.
   Chain rule, Theorem 4 (Section 5.3.1). Lesson: chain rule

4. Multiply through by $\sqrt{1-x^2}$

   Multiply every term by $\sqrt{1-x^2}$⁠, using $\sqrt{1-x^2}\cdot\sqrt{1-x^2} = 1 - x^2$⁠:

   $$(1-x^2)\,\frac{d^2y}{dx^2} - x\,\frac{dy}{dx} = 0.$$

   This is the required result, for every $x$ in $(-1, 1)$⁠.

   The target has $1 - x^2$ in front of $\frac{d^2y}{dx^2}$ and no fractions. Multiplying by the positive number $\sqrt{1-x^2}$ achieves exactly this, and since it is not zero it does not change which equations are true.
   $\left(\sqrt{p}\right)^2 = p$ for $p \ge 0$⁠.

5. The book's alternative: square first

   Write $y_1 = \frac{dy}{dx}$⁠, $y_2 = \frac{d^2y}{dx^2}$⁠. Squaring $\sqrt{1-x^2}\,y_1 = 1$ gives

   $$(1-x^2)\,y_1^2 = 1.$$

   Differentiate with the product rule, using $\frac{d}{dx}(y_1^2) = 2y_1y_2$ (chain rule):

   $$(1-x^2)\cdot 2y_1y_2 + y_1^2\,(0 - 2x) = 0.$$

   Divide by $2y_1$⁠, which is allowed because $y_1 = \frac{1}{\sqrt{1-x^2}} \gt 0$⁠:

   $$(1-x^2)\,y_2 - x\,y_1 = 0.$$
   Squaring removes the square root completely, so no chain rule on $\sqrt{1-x^2}$ is needed. The price is a division by $2y_1$ at the end, which the book does not comment on; it is valid because $y_1$ is never zero on $(-1,1)$⁠.
   $\frac{d}{dx}(1 - x^2) = 0 - 2x$⁠. Basics: derivative rules

6. Direct check: compute $\frac{d^2y}{dx^2}$ itself

   Write $\frac{dy}{dx} = (1-x^2)^{-\frac12}$ and use the chain rule:

   $$\begin{aligned} \frac{d^2y}{dx^2} &= -\frac12(1-x^2)^{-\frac32}\cdot(-2x) \\ &= \frac{x}{(1-x^2)^{\frac32}}. \end{aligned}$$

   Then

   $$(1-x^2)\cdot\frac{x}{(1-x^2)^{\frac32}} = \frac{x}{\sqrt{1-x^2}} = x\,\frac{dy}{dx},$$

   so $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$ once more.

   This brute-force route needs fractional powers but confirms the result independently. It also gives the actual second derivative, which the numerical check below uses.
   Power rule with $n = -\frac12$⁠, and $\frac{(1-x^2)^{1}}{(1-x^2)^{3/2}} = (1-x^2)^{-\frac12}$⁠.

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