Kerala KEAM 5 June 2024 Solved Paper

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). Quick Tip: 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.