Problem 17 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Solution of \(\left(\frac{x+y-1}{x+y-2}\right) \frac{d y}{d x}=\left(\frac{x+y+1}{x+y+2}\right)\), given that \(y=1\) when \(x=1\), is (A) \(\log \left|\frac{(x-y)^{2}-2}{2}\right|=2(x+y)\) (B) \(\log \left|\frac{(x-y)^{2}+2}{2}\right|=2(x-y)\) (C) \(\log \left|\frac{(x+y)^{2}+2}{2}\right|=2(x-y)\) (D) None of these

Step-by-Step Solution

Answer

Option D: None of these is correct.

Step 1: Separate Variables

Start by separating the variables in the differential equation given: \( \left(\frac{x+y-1}{x+y-2}\right) \frac{dy}{dx} = \left(\frac{x+y+1}{x+y+2}\right) \). Rearrange so that all terms involving \(y\) are on one side and the terms involving \(x\) are on the other. This is done by multiplying both sides by \(\frac{dx}{dy}\).

Step 2: Solve the Separated Equation

After separation, integrate both sides of the equation with respect to their respective variables. To keep track, let's denote the expression as: \( \int \frac{x+y-1}{x+y+2} dy = \int \frac{x+y+1}{x+y-2} dx \).

Step 3: Evaluate the Integrals

Evaluate the integrals obtained from separating the variables. These can become complex logarithmic functions depending on the simplification of each side, potentially involving integration by parts or substitution if needed.

Step 4: Apply Initial Condition

Once both integrals are solved, apply the given initial condition \( y=1 \) when \( x=1 \) to solve for any constant of integration. This will help in fully determining the particular solution to the differential equation.

Step 5: Match with Given Options

After applying the initial condition and obtaining the complete expression, compare the solution with the given options (A), (B), (C), and (D) to find the correct match.

Key Concepts

Variable Separation

Variable separation is a crucial technique for solving differential equations. It's like sorting out the terms so each variable has its side of the equation. This separation allows us to treat each variable separately in the integration process.
In the given problem, we started with a differential equation: \[ \left(\frac{x+y-1}{x+y-2}\right) \frac{dy}{dx} = \left(\frac{x+y+1}{x+y+2}\right) \] To separate the variables, we rearranged the equation, moving each variable to its respective side. This involved steps like multiplying both sides by \( \frac{dx}{dy} \) ensuring you end up with an expression ready for integration.

• Identify which terms belong to \( x \) and which belong to \( y \).
• Rearrange everything to group \( x \) and \( y \) on different sides of the equation.

Mastering variable separation is essential for tackling complex differential equations because it simplifies our work to the basic arithmetic of each variable.

Integration

Once the variables are separated, the next step is integration. Think of integration as finding a function whose derivative gives back the separated terms.
Using integration on both sides of the separated equation helps arrive at a function on each side. We focus on:\[ \int \frac{x+y-1}{x+y+2} dy = \int \frac{x+y+1}{x+y-2} dx \]With these integrals, it's all about finding antiderivatives. The complexity in these problems usually involves logarithms or substitutions, especially when dealing with fractions or more intricate expressions.

• Evaluate each side's integral independently.
• Understand that this step could involve advanced techniques like integration by parts.

Make sure to practice different integration methods. Building a varied toolkit makes solving these types of equations easier and more intuitive.

Initial Conditions

Initial conditions help define the particular solutions of differential equations. These are values provided to specify the constants that appear after integration.
For example, in the given problem, we used the information \( y=1 \) when \( x=1 \) to find any constant of integration. This step ensures that our solution isn't just a general solution, but a specific one that meets the given criteria of the problem.

• Substitute the initial conditions into your integrated result.
• Solve for the constant of integration, which aligns the equation to the specific initial values.

Using initial conditions is like setting the starting point of a graph. It ensures your solution fits exactly with the scenario depicted by the problem, making your answer meaningful and applicable.