Question

The function \(y = \tan^{-1}x\) satisfies differential equation

• A. \( (1+x^2)\frac{d^{n+2}y}{dx^{n+2}} + (2n+2)x\frac{d^{n+1}y}{dx^{n+1}} + n(n+1)\frac{d^ny}{dx^n} = 0 \)
• B. \( (1+x^2)\frac{d^{n+2}y}{dx^{n+2}} + (2n-1)x\frac{d^{n+1}y}{dx^{n+1}} + (n+1)\frac{d^ny}{dx^n} = 0 \)
• C. \( (1+x^2)\frac{d^{n+2}y}{dx^{n+2}} + (2n^2+n+1)\frac{d^{n+1}y}{dx^{n+1}} + (n-1)\frac{d^ny}{dx^n} = 0 \)
• D. \( (1+x^2)\frac{d^{n+2}y}{dx^{n+2}} + n(2n+1)\frac{d^{n+1}y}{dx^{n+1}} + n(n-1)\frac{d^ny}{dx^n} = 0 \)

Hint:

For problems involving finding the \(n^{th}\) order differential equation, the standard procedure is to differentiate once or twice until you can form an equation that is a polynomial in x multiplying derivatives of y. Then, apply Leibniz's theorem. The derivatives of the polynomial part (like \(1+x^2\) or \(2x\)) will terminate after a few terms, making the application straightforward.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires finding a general differential equation for the \(n^{th}\) derivative of \(y = \tan^{-1}x\). This is a typical application of Leibniz's theorem for the \(n^{th}\) derivative of a product of two functions.
Step 2: Key Formula or Approach:
1. Find the first few derivatives of \(y = \tan^{-1}x\) to get a relation involving a product. 2. Apply Leibniz's theorem to this relation. Leibniz's Theorem: \( D^n(uv) = \sum_{k=0}^n \binom{n}{k} D^{n-k}(u) D^k(v) \). In expanded form for the first few terms: \( D^n(uv) = u D^n(v) + n D(u) D^{n-1}(v) + \frac{n(n-1)}{2!} D^2(u) D^{n-2}(v) + \ldots \)
Step 3: Detailed Explanation:
Let \( y = \tan^{-1}x \). Differentiating once: \( y_1 = \frac{dy}{dx} = \frac{1}{1+x^2} \). Rearrange to form a product: \( (1+x^2)y_1 = 1 \). Differentiating this equation again with respect to x: Using the product rule: \( (1+x^2)y_2 + (2x)y_1 = 0 \). This is the differential equation for n=0. We now differentiate this equation \(n\) times using Leibniz's theorem. Let \(u = y_2\) and \(v = 1+x^2\). \( D^n[(1+x^2)y_2] + D^n[2xy_1] = 0 \) Term 1: \(D^n[(1+x^2)y_2]\) \[ (1+x^2)D^n(y_2) + n D(1+x^2)D^{n-1}(y_2) + \frac{n(n-1)}{2}D^2(1+x^2)D^{n-2}(y_2) + \ldots \] \[ (1+x^2)y_{n+2} + n(2x)y_{n+1} + \frac{n(n-1)}{2}(2)y_n \] \[ (1+x^2)y_{n+2} + 2nxy_{n+1} + n(n-1)y_n \] Term 2: \(D^n[2xy_1]\) Let \(u = y_1\) and \(v = 2x\). \[ 2[x D^n(y_1) + n D(x) D^{n-1}(y_1)] \] \[ 2[xy_{n+1} + n(1)y_n] = 2xy_{n+1} + 2ny_n \] Combine the two terms: \[ [(1+x^2)y_{n+2} + 2nxy_{n+1} + n(n-1)y_n] + [2xy_{n+1} + 2ny_n] = 0 \] Group like terms: \[ (1+x^2)y_{n+2} + (2nx + 2x)y_{n+1} + (n(n-1) + 2n)y_n = 0 \] \[ (1+x^2)y_{n+2} + (2n+2)xy_{n+1} + (n^2 - n + 2n)y_n = 0 \] \[ (1+x^2)y_{n+2} + (2n+2)xy_{n+1} + (n^2+n)y_n = 0 \] \[ (1+x^2)y_{n+2} + (2n+2)xy_{n+1} + n(n+1)y_n = 0 \] Step 4: Final Answer:
Replacing the subscript notation with the \(d/dx\) notation, we get the equation in option (A).