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

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that at first glance looks like a nightmare of calculus, but is actually a masterclass in algebraic elegance.

We are given the equation $y^{\frac{1}{4}}+y^{-\frac{1}{4}}=2x$ and asked to find the coefficients of a second-order differential equation. The secret to solving this isn't brute force differentiation; it is recognizing the hidden structure.

Look closely at the given equation:

If we set $t=y^{\frac{1}{4}}$, the equation becomes $t+\frac{1}{t}=2x$. Multiplying by $t$, we get $t^2+1=2xt$, or $t^2-2xt+1=0$.

This is a simple quadratic equation. Using the quadratic formula, we find:

We choose the positive root, so $y^{\frac{1}{4}}=x+\sqrt{x^2-1}$. We have successfully bridged the gap from a complex power equation to a manageable algebraic form.

Now, we need to find the differential equation. We could differentiate $y^{\frac{1}{4}}=x+\sqrt{x^2-1}$ directly, but that would be a path filled with messy chain rule terms.

Instead, let us differentiate both sides with respect to $x$:

Combining the right side, we get $\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}$. Notice that the numerator is exactly our original $y^{\frac{1}{4}}$.

Substituting this back, we get:

Multiplying by $4y^{\frac{3}{4}}$, we find $\frac{dy}{dx}=\frac{4y}{\sqrt{x^2-1}}$. This is much cleaner.

To get to the second derivative, let us cross-multiply to get $\sqrt{x^2-1}\frac{dy}{dx}=4y$. Now, square both sides to eliminate the radical:

Differentiate this with respect to $x$ using the product rule:

Dividing the entire equation by $2\frac{dy}{dx}$, we get:

Rearranging, we have $(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-16y=0$. Comparing this with the given $(x^2-1)\frac{d^{2}y}{d{x}^2}+\alpha x\frac{dy}{dx}+\beta y=0$, we identify $\alpha =1$ and $\beta =-16$.

The final calculation is:

And there you have it! A beautiful, logical journey from algebra to calculus, resulting in a clean, satisfying answer.