Miscellaneous · Q22: $y=e^{a\cos^{-1}x}$ satisfies $(1-x^2)y''-xy'-a^2y=0$

Basics you need

Solution

1. Where the derivatives exist

   $y$ is defined for $-1 \le x \le 1$⁠, but $\cos^{-1}x$ is differentiable only for $-1 < x < 1$⁠. So $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$⁠, and hence the relation to be shown, make sense for $-1 < x < 1$⁠. We work there.

   At $x = \pm 1$ the formula for $y'$ found below has $\sqrt{1-x^2} = 0$ in the denominator, so it cannot be used there; the definition decides. At $x = 1$⁠, $y(1) = e^0 = 1$⁠. Put $x = 1 - u$ with small $u > 0$ and $\theta = \cos^{-1}(1 - u)$⁠, so $\theta > 0$⁠, $\theta \to 0$ as $u \to 0^+$ ($\cos^{-1}$ is continuous, as remarked in Section 5.3.3), and $u = 1 - \cos\theta = 2\sin^2\frac{\theta}{2}$⁠. For $a \ne 0$ the left-hand difference quotient is $$\begin{aligned} \frac{y(1-u) - y(1)}{-u} &= -a\cdot\frac{e^{a\theta} - 1}{a\theta}\cdot\frac{\theta}{u}, \\ \frac{u}{\theta} &= \frac{\theta}{2}\left(\frac{\sin(\theta/2)}{\theta/2}\right)^2 \to 0. \end{aligned}$$ The middle factor tends to 1 (it is the difference quotient of $e^x$ at 0, and the derivative of $e^x$ at 0 is $e^0 = 1$⁠, Theorem 5), and $\frac{\sin t}{t} \to 1$ (Class XI, Limits and Derivatives), so $\frac{\theta}{u}$ grows without bound and so does the quotient: $y$ has no derivative at 1. With $\theta = \pi - \varphi$ and $\varphi$ small, the same computation works at $x = -1$⁠. The book writes $-1 \le x \le 1$ for the function; the relation itself holds on the open interval. (For $a = 0$⁠, $y = 1$ and everything is trivially 0.)
   Domain of the derivative of $\cos^{-1}$⁠: $(-1, 1)$ (Section 5.3.3; Basic 1). Lesson: derivatives of inverse trig functions

2. First derivative by the chain rule

   Outer function $e^u$⁠, inner function $u = a\cos^{-1}x$ with $\frac{du}{dx} = -\frac{a}{\sqrt{1-x^2}}$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= e^{a\cos^{-1}x}\cdot\left(-\frac{a}{\sqrt{1-x^2}}\right) \\ &= -\frac{ay}{\sqrt{1-x^2}}. \end{aligned}$$
   The derivative of $e^u$ with respect to $u$ is $e^u$ itself, which is $y$⁠. Multiplying by the inner derivative gives the result. Writing it in terms of $y$ keeps the next steps short.
   $\frac{d}{dx}e^x = e^x$ (Theorem 5, Section 5.4); chain rule (Theorem 4); Basic 1. Lesson: chain rule

3. Clear the square root

   Multiply both sides by $\sqrt{1-x^2}$ (positive for $-1 < x < 1$⁠):

   $$\sqrt{1-x^2}\,\frac{dy}{dx} = -ay. \qquad (1)$$
   Differentiating $-\frac{ay}{\sqrt{1-x^2}}$ directly would need the quotient rule and messy roots. Equation (1) has no fraction, so the product rule handles it cleanly. Example 38 uses the same idea.
   Multiplying both sides of an equation by the same non-zero quantity. Basics: solving for dy/dx

4. Differentiate equation (1) with the product rule

   Left side: $u = \sqrt{1-x^2}$ with $u' = \frac{-2x}{2\sqrt{1-x^2}} = -\frac{x}{\sqrt{1-x^2}}$⁠, and $v = y'$ with $v' = y''$⁠. Right side: $-a y'$⁠. So

   $$\begin{aligned} & \sqrt{1-x^2}\,\frac{d^2y}{dx^2} - \frac{x}{\sqrt{1-x^2}}\,\frac{dy}{dx} \\ &= -a\frac{dy}{dx}. \end{aligned}$$
   Equation (1) holds for every $x$ in $(-1, 1)$⁠, so both sides are the same function and we may differentiate both. The derivative of $\sqrt{1-x^2}$ uses the chain rule with inner function $1 - x^2$⁠.
   Product rule (rule (2), Section 5.3); $\frac{d}{dx}\sqrt{w} = \frac{1}{2\sqrt w}\frac{dw}{dx}$⁠; second order derivative (Section 5.7). Lesson: second order derivative

5. Multiply by $\sqrt{1-x^2}$ and use (1) again

   Multiplying every term by $\sqrt{1-x^2}$⁠:

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

   By (1), $\sqrt{1-x^2}\,\frac{dy}{dx} = -ay$⁠, so the right side is $-a(-ay) = a^2y$⁠:

   $$(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = a^2y,$$ $$(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0.$$
   Multiplying by $\sqrt{1-x^2}$ turns $\sqrt{1-x^2}\cdot\sqrt{1-x^2}$ into $1 - x^2$ and removes the fraction. The leftover $\sqrt{1-x^2}\,y'$ is exactly the left side of (1), which we replace by $-ay$⁠. Two minus signs give $+a^2y$⁠.
   Substituting a known relation, equation (1). Basics: solving for dy/dx

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