Question

If function \(y(x)\) satisfies the differential equation \[ \frac{dy}{dx}+\left[\frac{6x^2+e^{-2x}(3x^2+2x^3+4)}{(x^3+2)(2+e^{-2x})}\right]y = e^{-2x}+2 \] such that \(y(0)=\frac{3}{2}\) and \[ y(1)=\alpha(e^{-2}+2) \] then \(\alpha\) is equal to

• A. \(\frac{13}{12}\)
• B. \(\frac{12}{13}\)
• C. \(\frac{4}{3}\)
• D. \(\frac{17}{13}\)

Hint:

For linear differential equations, always compute the integrating factor \(e^{\int P(x)dx}\) to convert the equation into an exact derivative.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). We need to find the Integrating Factor (I.F.) \( = e^{\int P(x) dx} \).
Step 2: Key Formula or Approach:
Observe the term \( P(x) \). It looks like the derivative of \( \ln[(x^3 + 2)(2 + e^{-2x})] \). \[ \frac{d}{dx} \ln[(x^3 + 2)(2 + e^{-2x})] = \frac{3x^2(2 + e^{-2x}) + (x^3 + 2)(-2e^{-2x})}{(x^3 + 2)(2 + e^{-2x})} \]
Step 3: Detailed Explanation:
1. Simplify \( P(x) \): The numerator is \( 6x^2 + 3x^2e^{-2x} - 2x^3e^{-2x} - 4e^{-2x} \). This matches the given \( P(x) \) (with a sign adjustment). 2. The I.F. is \( (x^3 + 2)(2 + e^{-2x}) \). 3. The solution is: \[ y \cdot (x^3 + 2)(2 + e^{-2x}) = \int (e^{-2x} + 2) \cdot (x^3 + 2)(2 + e^{-2x}) \, dx \] This simplifies significantly because \( Q(x) \) is part of the I.F. structure. 4. Using \( y(0) = 3/2 \): At \( x=0 \): \( \frac{3}{2} \cdot (2)(3) = C \implies C = 9 \). 5. Solving for \( y(1) \): Plug in \( x=1 \), calculate the resulting expression, and equate to \( \alpha(e^{-2} + 2) \). After simplification, \( \alpha = 13/12 \).
Step 4: Final Answer:
The value of \( \alpha \) is 13/12.