Question

If $ f(x) = x - 1 $ and $ g(x) = e^x $, and the differential equation is: $ \frac{dy}{dx} = \left( e^{-2\sqrt{x}} g(f(f(f(x)))) - \frac{y}{\sqrt{x}} \right) $ with the initial condition $ y(0) = 0 $, then $ y(1) $ equals:

• A. \( \frac{2e - 1}{e^4} \)
• B. \( \frac{e - 1}{e^4} \)
• C. \( \frac{e^3 - 1}{e^4} \)
• D. \( \frac{e^2 - 1}{e^4} \)

Hint:

In solving differential equations involving exponential and square root terms, always simplify the given functions first, and use an integrating factor if needed.

Answer 1

The Correct Option is B

Given \( f(x) = x - 1 \) and \( g(x) = e^x \), and the differential equation \[\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g(f(f(f(x)))) - \frac{y}{\sqrt{x}} \right)\]We first determine \( f(f(f(x))) \). Since \( f(x) = x - 1 \), \( f(f(x)) = f(x - 1) = (x - 1) - 1 = x - 2 \), and \( f(f(f(x))) = f(x - 2) = (x - 2) - 1 = x - 3 \). Therefore, \( f(f(f(x))) = x - 3 \).
Substituting this into the differential equation yields \[\frac{dy}{dx} = e^{-2\sqrt{x}} e^{x - 3} - \frac{y}{\sqrt{x}}\]This simplifies to \[\frac{dy}{dx} = e^{-2\sqrt{x} + x - 3} - \frac{y}{\sqrt{x}}\]We now solve this differential equation with the initial condition \( y(0) = 0 \). Using an integrating factor, we find that \[y(1) = \frac{e - 1}{e^4}\]Hence, \( y(1) = \frac{e - 1}{e^4} \).