Question

Let $ f(x) = x - 1 $ and $ g(x) = e^x $ for $ x \in \mathbb{R} $. If $\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f\left(f(x)\right)\right) - \frac{y}{\sqrt{x}} \right), \, y(0) = 0,$ then $ y(1) $ is

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

Hint:

- For linear differential equations, use integrating factor method - Remember \( \int \frac{1}{\sqrt{x}} dx = 2\sqrt{x} + C \) - When applying initial conditions, solve for the constant immediately - Simplify exponential expressions carefully

Answer 1

The Correct Option is C

Step 1: Simplify the differential equation Given \( f(x) = x - 1 \) and \( g(x) = e^x \), we compute: \[ f(f(x)) = f(x - 1) = (x - 1) - 1 = x - 2 \] \[ g(f(f(x))) = e^{x - 2} \] Thus, the differential equation becomes: \[ \frac{dy}{dx} + \frac{y}{\sqrt{x}} = e^{-2\sqrt{x}} \cdot e^{x - 2} = e^{x - 2\sqrt{x} - 2} \]
Step 2: Solve using integrating factor The equation is linear of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where: \[ P(x) = \frac{1}{\sqrt{x}}, \quad Q(x) = e^{x - 2\sqrt{x} - 2} \] Integrating factor \( \mu(x) \): \[ \mu(x) = e^{\int P(x) dx} = e^{2\sqrt{x}} \] Multiply through by \( \mu(x) \): \[ e^{2\sqrt{x}} \frac{dy}{dx} + \frac{e^{2\sqrt{x}}}{\sqrt{x}} y = e^{x - 2} \] The left side is \( \frac{d}{dx}(y e^{2\sqrt{x}}) \), so: \[ \frac{d}{dx}(y e^{2\sqrt{x}}) = e^{x - 2} \] Integrate both sides: \[ y e^{2\sqrt{x}} = \int e^{x - 2} dx = e^{x - 2} + C \]
Step 3: Apply initial condition and find \( y(1) \) Using \( y(0) = 0 \): \[ 0 \cdot e^{0} = e^{-2} + C \Rightarrow C = -e^{-2} \] Thus: \[ y e^{2\sqrt{x}} = e^{x - 2} - e^{-2} \] At \( x = 1 \): \[ y(1) e^{2} = e^{-1} - e^{-2} = \frac{1}{e} - \frac{1}{e^2} \] \[ y(1) = \frac{e - 1}{e^3} \cdot \frac{1}{e} = \frac{e - 1}{e^4} \]