Question

If \( \frac{d^n y}{dx^n} = y_n \) and \( y = e^{\sqrt{x}} + e^{-\sqrt{x}} \), then find the value of \( 4x y_2 + 2 y_1 \).

• A. \( -y \)
• B. \( y \)
• C. \( 2y \)
• D. \( -2y \)

Hint:

In expressions involving \( e^{\sqrt{x}} \) and derivatives, common identities like \( 4x y'' + 2y' = y \) can save time.

Answer 1

The Correct Option is B

Given: \( y = e^{\sqrt{x}} + e^{-\sqrt{x}} \) Let \( u = \sqrt{x} \Rightarrow y = e^u + e^{-u} \Rightarrow \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) \[ \begin{align} \frac{dy}{dx} = (e^u - e^{-u}) \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}} \] Now compute: \[ y_1 = \frac{dy}{dx},\quad y_2 = \frac{d^2 y}{dx^2} \] Differentiating again: \[ \begin{align} y_2 = \frac{d}{dx} \left( \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}} \right) = \frac{(e^{\sqrt{x}} + e^{-\sqrt{x}}) \cdot \frac{1}{2\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - \left(e^{\sqrt{x}} - e^{-\sqrt{x}}\right) \cdot \frac{1}{2x\sqrt{x}} }{1} \] Tedious to simplify, but using identity: \[ 4x y_2 + 2y_1 = y \quad \text{(standard identity for this form)} \]