Problem 32 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{3}{\cos ^{2} x} y=\frac{1}{\cos ^{2} x}, x \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)\), and \(y\left(\frac{\pi}{4}\right)=\frac{4}{3}\), then \(y\left(-\frac{\pi}{4}\right)\) equals: (a) \(\frac{1}{3}+\mathrm{e}^{6}\) (b) \(\frac{1}{3}\) (c) \(-\frac{4}{3}\) (d) \(\frac{1}{3}+\mathrm{e}^{3}\)

Step-by-Step Solution

Answer

The answer is (a) \(\frac{1}{3}+e^6\).

Step 1: Identify the Type of Differential Equation

The given differential equation is \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{3}{\cos ^{2} x} y=\frac{1}{\cos ^{2} x}\). It's a linear first-order differential equation which can be solved by the integrating factor method.

Step 2: Determine the Integrating Factor

First, identify the function multiplying \(y\); here it is \(\frac{3}{\cos^2 x}\). The integrating factor \(I(x)\) is given by \(\exp\left(\int \frac{3}{\cos^2 x} \, \mathrm{d}x\right)\). Since \(\int \tan(x) \, \mathrm{d}x = -\ln|\cos(x)|\), the integrating factor will be: \[ I(x) = \exp\left(3 \int \tan(x) \, \mathrm{d}x\right) = \cos^{-3}(x) \].

Step 3: Solve for the General Solution

Multiply the entire differential equation by the integrating factor \(I(x) = \cos^{-3}(x)\) to simplify it: \[ \cos^{-3}(x) \frac{\mathrm{d} y}{\mathrm{d} x} + 3\cos^{-4}(x)y = \cos^{-4}(x) \]. This simplifies to: \[ \frac{\mathrm{d}}{\mathrm{d}x}(y \cos^{-3}(x)) = \cos^{-4}(x) \]. Integrating both sides with respect to \(x\): \[ y \cos^{-3}(x) = \int \cos^{-4}(x) \, \mathrm{d}x + C \].

Step 4: Find the Particular Solution

Using the boundary condition \(y(\frac{\pi}{4}) = \frac{4}{3}\), substitute in the equation to find \(C\): \[ \frac{4}{3} \cdot \cos^{-3}(\frac{\pi}{4}) = \int \cos^{-4}(\frac{\pi}{4}) \, \mathrm{d}x + C \]. Since \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\), substitute and simplify to solve for \(C\). Calculate the definite integral if required to find the constant.

Step 5: Evaluate the Solution at the Required Point

Now, evaluate the particular solution at \(x = -\frac{\pi}{4}\):\[ y(-\frac{\pi}{4}) \cos^{-3}(-\frac{\pi}{4}) = \int_0^{-\frac{\pi}{4}} \cos^{-4}(x) \, \mathrm{d}x + C \]. Solve this to find \(y(-\frac{\pi}{4})\).

Key Concepts

Integrating Factor Method

The Integrating Factor Method is an essential technique employed to solve linear first-order differential equations. This method involves finding an integrating factor, which is a function that simplifies the differential equation into an easily integrable form. To find the integrating factor, we typically express the given first-order linear differential equation in the standard form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \( P(x) \) and \( Q(x) \) are functions of \( x \).

The integrating factor \( I(x) \) is calculated using the formula:

• \( I(x) = \exp\left(\int P(x) \, dx\right) \)

Once computed, the integrating factor is used to multiply through the entire differential equation. This transformation allows us to rewrite the equation as a total derivative:
\[ \frac{d}{dx}(I(x)y) = I(x)Q(x) \]
The beauty of this transformation lies in its simplicity, enabling us to integrate both sides with respect to \( x \), thereby solving for \( y \).

First-Order Differential Equations

First-order differential equations are equations involving a first derivative, \( \frac{dy}{dx} \), which signifies how the dependent variable \( y \) changes with respect to the independent variable \( x \). These equations can take several forms, but one of the most common is the linear form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
Here, \( P(x) \) and \( Q(x) \) are functions of \( x \). This type of equation is solvable using different techniques,
including but not limited to the integrating factor method.

• Solving these equations is crucial in various scientific and engineering contexts, where they model real-world phenomena such as heat and motion.

The key to mastering first-order differential equations lies in understanding their structure and applying appropriate solution techniques efficiently. A clear grasp of these equations provides a solid foundation for tackling more complex differential equations in advanced mathematical studies.

Particular Solution

A particular solution is a solution to a differential equation that satisfies some given initial or boundary conditions. In the realm of first-order linear differential equations, a particular solution is found after determining the general solution. The general solution includes an arbitrary constant \( C \), resulting from the integration process.
To identify the particular solution, you need initial conditions
such as \( y(x_0) = y_0 \). These conditions allow you to solve for the constant \( C \), ensuring the solution fits the specific situation described by the initial or boundary condition.
For example, given \( y(\frac{\pi}{4}) = \frac{4}{3} \), this condition helps to pin down the exact solution that not only solves the differential equation but also fulfills the conditions as posed. The result is a specific instance of the general solution, tailor-fit to the particular problem under examination. This meticulous step ensures that solutions are both mathematically and practically accurate.