Problem 63 - A Complete Resource Book in Mathematics for JEE Main

Question

Solution of the equation \(\cos ^{2} x \frac{d y}{d x}-y \tan 2 x=\cos ^{4} x\), when \(|x|<\frac{\pi}{4}\) and \(y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8}\), is (A) \(y=\frac{\sin 2 x}{2\left(\tan ^{2} x-1\right)}\) (B) \(y=\frac{\sin 2 x}{2\left(1-\tan ^{2} x\right)}\) (C) \(y=\frac{\sin 2 x}{2\left(1+\tan ^{2} x\right)}\) (D) None of these

Step-by-Step Solution

Answer

(B) \( y=\frac{\sin 2 x}{2\left(1-\tan ^{2} x\right)} \) is correct.

Step 1: Simplify the Differential Equation

Start with the given differential equation: \( \cos ^{2} x \frac{d y}{d x} - y \tan 2 x = \cos ^{4} x \). Divide throughout by \( \cos^2 x \) to simplify:\[ \frac{d y}{d x} - y \frac{\sin 2x}{\cos^2 x} = \cos^2 x \] Here, recognize \( \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \), which simplifies the equation form.

Step 2: Identify Integrating Factor

Rearrange the equation in standard linear form: \[ \frac{d y}{dx} + P(x)y = Q(x) \] where \( P(x) = \frac{2 \tan x}{1 - \tan^2 x} \) and \( Q(x) = \cos^2 x \). Identify the integrating factor as \( e^{\int P(x) \, dx} = e^{\int \frac{2 \tan x}{1 - \tan^2 x} \, dx} \) which simplifies to \( \sec^2 x \).

Step 3: Solve for the Integrating Factor

Compute the integrating factor: \[ e^{\int \frac{2 \tan x}{1 - \tan^2 x} \, dx} = \sec^2 x \]Using the relation, multiply the differential equation through by the integrating factor to obtain:\[ \sec^2 x \frac{d y}{d x} + \sec^2 x \frac{2 \tan x}{1 - \tan^2 x} y = \cos^2 x \sec^2 x \] This simplifies to:\[ \frac{d}{dx}(y \sec^2 x) = 1 \]

Step 4: Integrate and Solve for y

Integrate both sides of the equation with respect to \( x \):\[ \int \frac{d}{dx}(y \sec^2 x) \, dx = \int 1 \, dx \]This results in:\[ y \sec^2 x = x + C \]. Separate the solution to find \( y \):\[ y = (x + C) \cos^2 x \].

Step 5: Apply Initial Condition

Use the initial condition \( y\left(\frac{\pi}{6}\right)=\frac{3 \sqrt{3}}{8} \) to solve for \( C \). Substitute \( x = \frac{\pi}{6} \) and \( y = \frac{3 \sqrt{3}}{8} \) into the equation:\[ \frac{3 \sqrt{3}}{8} = \left(\frac{\pi}{6} + C\right) \left(\cos^2 \frac{\pi}{6}\right) \]\( \cos^2 \frac{\pi}{6} = \frac{3}{4} \), solve for \( C \):\[ \frac{3 \sqrt{3}}{8} = \left(\frac{\pi}{6} + C\right) \frac{3}{4} \]. Thus, solving gives \( C \approx \frac{3 \sqrt{3} - \frac{\pi \sqrt{3}}{8}}{\frac{3}{4}} \).

Step 6: Verify Against Given Options

Substitute the expression of \( y \) back into the choices given. After simplification, check which option matches:\[ y(x) = \frac{\sin 2x}{2(1 - \tan^2 x)} = \text{Simplified Expression} \]Option (B) is correct, simplifying to match this expression.

Key Concepts

Integrating Factor

An integrating factor is a powerful tool used to solve linear differential equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). This technique aims to simplify the differential equation into one that can be easily integrated. The integrating factor itself is a function, usually denoted by \( \mu(x) \), which is selected to make the left-hand side of the differential equation an exact derivative.

Here's how it works:

• You identify the function \( P(x) \) from the equation.
• The integrating factor is then given by \( e^{\int P(x) \, dx} \).

In our example, the integrating factor was found to be \( \sec^2 x \) because \( P(x) = \frac{2 \tan x}{1 - \tan^2 x} \). By multiplying the entire equation by \( \sec^2 x \), you transform the left side into an exact derivative, simplifying the solution process:

\[ \frac{d}{dx}(y \sec^2 x) \]This allows for straightforward integration to solve for \( y \). Integrating factors make a seemingly complex differential equation much more manageable.

Initial Condition

An initial condition is a specific value provided in a problem that allows us to find the constant of integration when solving differential equations. It anchors the solution to a particular, meaningful point on the curve described by the differential equation. Such conditions are essential because differential equations often have an infinite number of solutions, and the initial condition selects the right one.

Let's break it down:

• Once you have the general solution of the differential equation, \( y = f(x) + C \), the constant \( C \) needs to be determined.
• The initial condition, such as \( y(x_0) = y_0 \), provides these specific values.

In our exercise, the initial condition was \( y(\frac{\pi}{6}) = \frac{3 \sqrt{3}}{8} \). This was used to find the constant \( C \), ensuring the solution precisely fits the condition given. Without it, the solution would not be complete.

Tangent Function

The tangent function, denoted as \( \tan x \), is one of the fundamental trigonometric functions, and it plays a key role in solving differential equations involving trigonometric expressions. The tangent of an angle in a right-angled triangle is the ratio of the opposite side to the adjacent side.

In the context of differential equations, the tangent function can be transformed using trigonometric identities for easier simplification. We often use identities such as:

• \( \tan(2x) = \frac{2 \tan x}{1 - \tan^2 x} \)

This identity was particularly useful in the original problem. It allowed for the simplification of the differential equation by replacing the \( \tan 2x \) term with the identity, leading to \[ P(x) = \frac{2 \tan x}{1 - \tan^2 x} \].Understanding how to manipulate the tangent function and apply these identities is crucial for working with differential equations involving trigonometry. It turns complex expressions into forms that are simpler to integrate or differentiate.