Question

The integrating factor of

\[ (1 + 2e^{-x}) \frac{dy}{dx} - 2e^{-x} y = 1 + e^{-x} \] is:

• A. \( 2e^{-x} \)
• B. \( 1 + e^{-x} \)
• C. \( 1 - e^{-x} \)
• D. \( 1 - 2e^{-x} \)
• E. \( 1 + 2e^{-x} \)

Hint:

In linear first-order differential equations, the integrating factor is given by \( \mu(x) = e^{\int P(x) \, dx} \), where \( P(x) \) is the coefficient of \( y \). This factor helps simplify the equation into an exact differential equation.

Answer 1

Answer: (E)

The given differential equation is: \[ (1 + 2e^{-x}) \frac{dy}{dx} - 2e^{-x} y = 1 + e^{-x} \] This is a linear first-order differential equation of the form: \[ \frac{dy}{dx} + P(x) y = Q(x) \] To solve it, we need to find the integrating factor. The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} \] We need to rewrite the given equation in the standard form: \[ \frac{dy}{dx} + \left( \frac{-2e^{-x}}{1 + 2e^{-x}} \right) y = \frac{1 + e^{-x}}{1 + 2e^{-x}} \] Thus, the integrating factor \( \mu(x) \) is: \[ \mu(x) = e^{\int \frac{-2e^{-x}}{1 + 2e^{-x}} \, dx} \] To simplify this integral, we observe that the factor \( 1 + 2e^{-x} \) makes the expression simpler to integrate, and we find that: \[ \mu(x) = 1 + 2e^{-x} \] Thus, the integrating factor is \( 1 + 2e^{-x} \), which corresponds to option (E).