The solution of initial value problem \(\mathrm{x}(\mathrm{dy} / \mathrm{dx})=\mathrm{x}+\mathrm{y} ; \mathrm{y}(1)\) \(=1\) is \(\mathrm{y}=\) (A) \(x \log \overline{x-1}\) (B) \(x \log x+1\) (C) \(x(\log x+1)\) (D) none of these

The integrating factor method is a crucial technique for solving first-order linear differential equations, a key topic for students preparing for JEE Maths and other competitive exams.

The method starts by identifying a differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \), and \( Q(x) \), are functions of \( x \). The goal is to multiply each term in the differential equation by an 'integrating factor' that will allow the left-hand side to be written as the derivative of a product of two functions. This integrating factor is generally denoted as \( e^{\int P(x) dx} \).

Here's a brief overview of how to use the integrating factor method:

• Identify \( P(x) \) from the original equation and compute the integrating factor \( e^{\int P(x) dx} \).
• Multiply the entire differential equation by the integrating factor to get an exact differential on the left side.
• Integrate both sides of the equation with respect to \( x \) to find \( y \).
• Using the initial conditions provided to solve for any constants of integration.

When solving the example problem, \( x (\frac{dy}{dx}) = x + y \), we identified \( P(x) = \frac{1}{x} \) and found the integrating factor to be \( x \). After applying the integrating factor, we integrated to find the value of \( y \) and used the given initial condition to solve for the constant of integration.

In the context of differential equations, an initial value problem (IVP) is one where we are given a differential equation along with the value of the unknown function at a specific point. This additional information allows us to find a specific solution out of the infinite family of potential solutions that satisfy the given differential equation.

The solution to an initial value problem typically involves these steps:

• Integrate the differential equation to get a general solution that includes a constant of integration.
• Apply the given initial condition, which is a point on the function, to calculate the constant.
• Substitute the constant back into the general solution to obtain the particular solution that satisfies both the differential equation and the initial condition.

In the given exercise, we used the initial condition \( y(1) = 1 \) after integrating to determine the constant of integration \( C \). It is through this method that we found the particular solution that uniquely corresponds to the initial state provided, which in this case did not align with any of the provided answer choices.

Differential equations are a fundamental topic in the JEE Mathematics syllabus, which require a strong understanding for anyone preparing for the JEE (Joint Entrance Examination). They offer a way to describe changing quantities mathematically and appear in various forms with each having a different solution method.

In JEE Maths, students encounter:

• First-Order Equations: Including separable, homogeneous, and linear types.
• Higher-Order Equations: Linear equations with constant or variable coefficients.
• Applications: Modeling real-life situations like radioactive decay, population dynamics, and thermal problems.

Strong skills in solving differential equations can significantly benefit students in JEE Maths, as they often come across in both the problems and the solutions. The integrating factor method, showcased in our example problem, is just one of the many techniques that students must master. With continuous practice and a solid grasp of concepts like IVPs and the integrating factor method, students can enhance their problem-solving abilities and increase their chances of success in competitive exams like the JEE.