Problem 14 - A Complete Resource Book in Mathematics for JEE Main

Question

The solution of the differential equation \((x-y)(2 d y-d x)=3 d x-5 d y\) is (A) \(2 x-y=\log (x-y+z)+c\) (B) \(2 x+y=\log (x-y+z)+c\) (C) \(2 y-x=\log (x-y+z)+c\) (D) None of these

Step-by-Step Solution

Answer

(C) \(2y-x=\log(x-y+z)+c\)

Step 1: Simplify the Differential Equation

The given differential equation is \((x-y)(2 dy - dx) = 3 dx - 5 dy\). Distributing \((x-y)\) gives \(2(x-y)dy - (x-y)dx = 3dx - 5dy\). Rearrange all the terms such that all \(dy\) terms are on one side and all \(dx\) terms on the other. This results in \(2(x-y+z)dy = (x-y+3)dx\). Divide the equation through by \(dx\) to obtain \(2 \frac{dy}{dx} = \frac{x-y+3}{x-y+z}\).

Step 2: Separate Variables

Rearrange the equation from Step 1 to separate the variables \(y\) and \(x\): \(2 dy = \frac{x-y+3}{x-y+z} dx\). This step makes it suitable for integration by making each variable's terms separate from each other.

Step 3: Integrate Both Sides

Integrate both sides of the equation from Step 2. Integrating the left side with respect to \(y\) and the right side with respect to \(x\) gives \(2y = \log(x-y+z) + 2c\). The constant of integration is simplified to a general constant \(c\).

Step 4: Rearrange the Integrated Result

Rearrange the integrated equation to match one of the given options. We have \(2y - x = \log(x-y+z) + c\). This matches with option (C).

Key Concepts

Integration Techniques

Integration techniques are crucial to solving differential equations like the one in our exercise. The act of integration allows us to find the antiderivatives or to determine the function that a given derivative comes from. When tackling a differential equation, identifying the correct technique and applying it efficiently is key.

• **Basic Integration:** This technique involves finding the antiderivative of standard functions like polynomials, exponential functions, and trigonometric functions.
• **Substitution:** Useful when the integration involves a composite function, substitution helps simplify the integration process by introducing a new variable.
• **Partial Fractions:** Helpful for integrating rational functions; this technique resolves a complex fraction into simpler parts that are easier to integrate.

In our given differential equation, after separating the variables, we integrated each side independently. The left side with respect to \(y\) and the right with respect to \(x\). The outcome leads directly to the solution, integrating both constant and variable components and combining them with the constant of integration to find the complete, general solution. Remember, practice with a variety of equations is the best method to master these integration techniques.
Understanding when and how to apply each technique is essential, helping you navigate through even the trickiest of equations.

Variable Separation

Variable separation is a powerful algebraic technique used in differential equations that allows for simplification by isolating variables on different sides of the equation. This technique is perfectly suited for first-order differential equations like the one in our example.

• **Understanding the Structure:** Start by identifying terms involving each variable. The goal is to rearrange the equation such that all terms involving one variable are on one side, and the terms involving the other variable are on the opposite side.
• **Manipulation:** This often involves distributing factors, combining like terms, and sometimes factoring to ensure each term involving \(dy/dx\) is only linked to its respective variable.
• **Balancing the Equation:** After separation, ensure the equation is balanced such that each side can independently transform through integration.

In our exercise, we began with the equation \(2(x-y+z)dy = (x-y+3)dx\). Through careful balancing and manipulation, this was rearranged to separate \(dy\) and \(dx\) into their respective sides, which sets the stage for integration. The clarity this separation brings is invaluable, simplifying what initially appears complex into a straightforward format for solving.

First Order Differential Equations

First-order differential equations are equations involving the first derivative of a function. These equations express a relationship between a function and its rate of change. Such equations are foundational in modeling various real-world situations such as population growth, radioactive decay, or cooling of objects.

• **Standard Form:** A typical first-order differential equation is represented as \(dy/dx = f(x,y)\), where \(f(x,y)\) is a function involving both \(x\) and \(y\).
• **Solution Characteristics:** The solutions to first-order differential equations can be functions that describe a variety of behaviors, from simple linear change to more complex patterns.
• **Methods of Solution:** Techniques such as separation of variables, as demonstrated in our example, are particularly effective. Other methods like integrating factors or exact equations might be used depending on the specific form.

The equation in our exercise, after some rearrangement and simplification, revealed its true characteristic as a first-order equation, which allowed us to apply separation of variables and integration. Understanding the basic character of these equations helps predict the behavior of their solutions and informs the choice of methods to employ in finding a solution.