Problem 24 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(y=y(x)\) and \(\frac{2+\sin x}{y+1}\left(\frac{d y}{d x}\right)=-\cos x, y(0)=1\), then \(y\left(\frac{\pi}{2}\right)=\) (A) \(\frac{1}{3}\) (B) \(\frac{2}{3}\) (C) \(-\frac{1}{3}\) (D) 1

Step-by-Step Solution

Answer

The value of \( y\left(\frac{\pi}{2}\right) \) is \( \frac{1}{3} \). Therefore, the answer is (A).

Step 1: Simplify the Differential Equation

The given differential equation is \( \frac{2 + \sin x}{y + 1} \frac{dy}{dx} = -\cos x \). We simplify this to get: \( \frac{dy}{dx} = \frac{-(y + 1) \cos x}{2 + \sin x} \). This will be useful for separation of variables.

Step 2: Separate the Variables

Rearrange the terms to separate the variables: \( \frac{dy}{y + 1} = \frac{-\cos x}{2 + \sin x} dx \). This sets up the equation for integration with \( y \) on one side and \( x \) on the other.

Step 3: Integrate Both Sides

Integrate both sides to find \( y \):- The left side: \( \int \frac{dy}{y + 1} = \ln |y + 1| + C_1 \).- The right side: Let's use substitution, let \( u = 2 + \sin x \), then \( du = \cos x \, dx \), which simplifies the integration to \( \int \frac{-1}{u} du = -\ln |u| + C_2 \).

Step 4: Solve the Integral Equation

Combine the integrated results:\( \ln |y + 1| = -\ln |2 + \sin x| + C \).This becomes \( |y + 1| = \frac{C}{2 + \sin x} \).

Step 5: Use Initial Condition

Given \( y(0) = 1 \), substitute \( x = 0 \) and \( y = 1 \) into the equation: \( |1 + 1| = \frac{C}{2 + \sin 0} \) leading to \( 2 = \frac{C}{2} \). Thus, \( C = 4 \).

Step 6: Solve for y(x)

Substitute \( C = 4 \) back into the solution:\( |y + 1| = \frac{4}{2 + \sin x} \).Calculate for \( y(\frac{\pi}{2}) \):\[ y + 1 = \frac{4}{2 + 1} = \frac{4}{3} \]Thus, \( y = \frac{4}{3} - 1 = \frac{1}{3} \).

Step 7: Choose the Correct Answer

Comparing with the multiple choice answers, the value of \( y(\frac{\pi}{2}) \) is \( \frac{1}{3} \), which corresponds to option (A).

Key Concepts

Separation of Variables

The technique of separation of variables is a powerful method for solving ordinary differential equations (ODEs). This method allows us to rearrange a differential equation so that each variable and its differential are on opposite sides of the equation. In simpler terms, you put all the terms involving 'y' with 'dy' on one side, and all the terms involving 'x' with 'dx' on the other side.

Consider the example equation given in the exercise:

• The initial equation was \( \frac{2 + \sin x}{y + 1} \frac{dy}{dx} = -\cos x \)
• We rearranged it as \( \frac{dy}{y + 1} = \frac{-\cos x}{2 + \sin x} dx \)

This pivotal step aligns our equation for efficient integration, helping us move towards finding the specific solution for 'y' with respect to 'x'. Applying separation of variables helps streamline the process of solving differential equations by simplifying the equation into manageable integrals.

Initial Value Problems

Initial value problems (IVPs) are a class of differential equations with given specific values at a particular point. These serve as conditions to identify the unique solution to a differential equation. The given exercise provides an initial value condition:

• \( y(0) = 1 \)

This means that when \( x = 0 \), the function \( y \) equals 1. Using this condition is crucial because once the differential equation is solved, it helps us find the constant of integration, denoted here as 'C'.

In step 5 of the solution, by substituting \( x = 0 \) and \( y = 1 \) in the integrated equation, we've obtained

• \( |1 + 1| = \frac{C}{2 + 0} \), resulting in \( C = 4 \)

This result is then used to develop the complete solution of the equation, ensuring the computed solution satisfies the initial condition.

Integration Techniques

Integration techniques are mathematical methods used to solve integrals. In this exercise, various integration techniques are applied to evaluate the integrals after the separation of variables step. Let’s break it down:

• For the left-hand side: We integrated \( \int \frac{dy}{y + 1} \), which yields \( \ln |y + 1| \)
• For the right-hand side: A substitution method was used. By setting \( u = 2 + \sin x \), with \( du = \cos x \, dx \), the integral becomes \( \int \frac{-1}{u} du \), leading to \(-\ln |u| \)

Each of these steps simplifies describing the rate of change, allowing us to effectively integrate both sides of the equation. Successfully integrating with these techniques helps unlock the path to a solution by transforming complex differential equations into solvable formats. After integration, combining both sides provides an equation where the initial conditions can be applied to extract precise values for any constants involved.