WBJEE 2024 Math Paper

Let's solve the given differential equation and find the limit $\lim\limits_{x \to 1} y(x)$.The differential equation given is:$x y' + y - e^x = 0$ This is a first-order linear differential equation. We can rewrite it in the form:$y' + \;\frac{1}{x} y = \;\frac{e^x}{x}$We will solve this using the integrating factor method. The integrating factor (IF) is given by:$\mu(x) = e^{\int\;\frac{1}{x} dx} = e^{\ln |x|} = |x|$ Since we are not given a specific interval for $x$, we assume $x$ is positive:$\mu(x) = x$Multiplying both sides of the differential equation by the integrating factor:$\& x y' + y = e^x \Rightarrow x y' + y = e^x \Rightarrow x \left(y' + \;\frac{1}{x} y\right) = e^x$$\& x \cdot y' + y = e^x$ Recognize that the left side is the derivative of $x y$ :$\;\frac{d}{dx} (x y) = e^x$Integrate both sides with respect to x :$x y = \int e^x dx = e^x + C$So we have:$y = \;\frac{e^x + C}{x}$ Apply the initial condition $y(a) = b$ :$\& b = \;\frac{e^a + C}{a}$$\& C = a b - e^a$ Thus, the solution to the differential equation is:$y = \;\frac{e^x + a b - e^a}{x}$Now, we need to determine:$\lim\limits_{x \to 1} y(x)$ Substituting $x = 1$ into the solution:$y(1) = \;\frac{e + a b - e^a}{1} = e + a b - e^a$Therefore:The correct answer is Option D$e + a b - e^a$