Question

If $y^{1/4} + y^{-1/4} = 2x$, and $(x^2 - 1)\frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$, then $|\alpha - \beta|$ is equal to _________.

Hint:

When you see an expression like $y^{1/n} + y^{-1/n} = 2x$, it often leads to a solution of the form $y = (x + \sqrt{x^2-1})^n$. This is a standard substitution in differential calculus.

Answer 1

Correct Answer: 17

Step 1: Understanding the Concept:
This problem involves differentiating a function defined implicitly and manipulating the resulting differential equation to match a given form.
Step 2: Detailed Explanation:
Given $y^{1/4} + y^{-1/4} = 2x$.
Let $y^{1/4} = t$, then $t + \frac{1}{t} = 2x \implies t^2 - 2xt + 1 = 0$.
$t = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}$.
So, $y^{1/4} = x + \sqrt{x^2 - 1} \implies y = (x + \sqrt{x^2 - 1})^4$.
Differentiating with respect to $x$:
\[ \frac{dy}{dx} = 4(x + \sqrt{x^2 - 1})^3 \left( 1 + \frac{x}{\sqrt{x^2 - 1}} \right) = \frac{4(x + \sqrt{x^2 - 1})^4}{\sqrt{x^2 - 1}} = \frac{4y}{\sqrt{x^2 - 1}} \]
So, $\sqrt{x^2 - 1} \frac{dy}{dx} = 4y$.
Squaring both sides: $(x^2 - 1) \left(\frac{dy}{dx}\right)^2 = 16y^2$.
Differentiating again with respect to $x$:
\[ (x^2 - 1) \cdot 2 \frac{dy}{dx} \frac{d^2y}{dx^2} + 2x \left(\frac{dy}{dx}\right)^2 = 32y \frac{dy}{dx} \]
Dividing by $2 \frac{dy}{dx}$ (assuming $y'$ is not zero):
\[ (x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = 16y \implies (x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} - 16y = 0 \]
Comparing with $(x^2 - 1)y'' + \alpha x y' + \beta y = 0$:
$\alpha = 1$ and $\beta = -16$.
$|\alpha - \beta| = |1 - (-16)| = 17$.
Step 3: Final Answer:
The value of $|\alpha - \beta|$ is 17.