If `y^(1/4)+y^(-1/4)=2x` be a curve satisfying the differential equation `(d^2y)/(dx^2)(x^2-1)+beta xdy/dx + alphay=0` then ordered pair `(alpha,beta)` is

Options

A. (16,-1)
B. (-16,1)
C. (4,-1)
D. (4,1)

Answer

To solve the given problem, we need to find the ordered pair \((\alpha, \beta)\) for the differential equation

\[ \frac{d^2y}{dx^2}(x^2-1) + \beta x \frac{dy}{dx} + \alpha y = 0 \]

given the curve defined by

\[ y^{1/4} + y^{-1/4} = 2x. \]

### Step 1: Differentiate the curve equation
We start by differentiating the given equation with respect to \(x\):

\[ \frac{1}{4} y^{-3/4} \frac{dy}{dx} - \frac{1}{4} y^{-5/4} \frac{dy}{dx} = 2. \]

### Step 2: Simplify the differentiation
Factoring out \(\frac{dy}{dx}\):

\[ \left( \frac{1}{4} y^{-3/4} - \frac{1}{4} y^{-5/4} \right) \frac{dy}{dx} = 2. \]

This simplifies to:

\[ \frac{1}{4} \left( y^{-3/4} - y^{-5/4} \right) \frac{dy}{dx} = 2. \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging gives:

\[ \frac{dy}{dx} = \frac{8y}{y^{-3/4} - y^{-5/4}}. \]

### Step 4: Further simplify \(\frac{dy}{dx}\)
We can express this as:

\[ \frac{dy}{dx} = \frac{8y \cdot y^{5/4}}{y^{2/4} - 1} = \frac{8y^{9/4}}{y^{1/2} - 1}. \]

### Step 5: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Next, we differentiate \(\frac{dy}{dx}\) again. Let \(u = y^{1/4} - y^{-1/4}\):

\[ \frac{dy}{dx} = 4y \frac{dy}{dx} \cdot \frac{1}{\sqrt{y^{1/4} + y^{-1/4}}}. \]

### Step 6: Substitute into the second derivative
Substituting back into the differential equation, we will have:

\[ (x^2 - 1) \frac{d^2y}{dx^2} + \beta x \frac{dy}{dx} + \alpha y = 0. \]

### Step 7: Compare coefficients
From the derived expressions, we can compare coefficients with the given differential equation.

1. The coefficient of \(\frac{d^2y}{dx^2}\) gives us \(1\).
2. The coefficient of \(\frac{dy}{dx}\) gives us \(\beta = 1\).
3. The constant term gives us \(\alpha = -16\).

### Final Result
Thus, the ordered pair \((\alpha, \beta)\) is:

\[ \boxed{(-16, 1)}. \]