Question

The equation of the curve passing through \([1,(\pi / 4)]\) and having the slope \([(\sin 2 \mathrm{y}) /(\mathrm{x}+\tan \mathrm{y})]\) at \((\mathrm{x}, \mathrm{y})\) is: (A) \(x=\tan y\) (B) \(\mathrm{y}=2 \tan \mathrm{x}\) (C) \(y=\tan x\) (D) \(x=2 \tan y\)

Short answer

The correct answer is (D): \(x = 2 \tan y\).

Step-by-step solution

Write down the differential equation

We are given the slope of the curve at any point \((x, y)\) as \[\frac{dy}{dx} = \frac{\sin{2y}}{x + \tan{y}}.\]

Solve the differential equation

We can rewrite the differential equation as \[(x + \tan{y}) \: dy = (\sin{2y}) \: dx.\] Now integrate both sides with respect to their respective variables: \[\int (x + \tan{y}) \: dy = \int \sin{2y} \: dx.\]

Integration

Integrating both sides, we get \[xy + \int \tan{y} \: dy = x \sin{2y} + C,\] where \(C\) is the constant of integration. To find the integral of \(\tan{y}\), we use the formula \[\int \tan{y} \: dy = \ln |\sec y|.\] So, our equation becomes \[xy + \ln |\sec y| = x \sin{2y} + C.\]

Apply the initial condition of passing through \((1, \frac{\pi}{4})\)

The curve passes through the point \(\left(1, \frac{\pi}{4}\right)\), so when \(x = 1\) and \(y = \frac{\pi}{4}\), the equation must hold true. Plugging these values in the equation, we get \[\frac{\pi}{4} + \ln \left|\sec \left(\frac{\pi}{4}\right)\right| = \sin \frac{\pi}{2} + C.\] Since \(\sin \frac{\pi}{2} = 1\) and \(\sec \frac{\pi}{4} = \sqrt{2}\), we get \[\frac{\pi}{4} + \ln \sqrt{2} = 1 + C.\] Solving this equation for the constant \(C\), we find \[C = 1 - \frac{\pi}{4} - \ln \sqrt{2}.\]

Find the correct choice

Our final equation for the curve is \[xy + \ln |\sec y| = x \sin{2y} + 1 - \frac{\pi}{4} - \ln \sqrt{2}.\] Now, we will compare this equation with each given option and see which option satisfies this equation: (A) \(x = \tan y\): This option doesn't satisfy our equation. (B) \(y = 2 \tan x\): This option also doesn't satisfy our equation. (C) \(y = \tan x\): This can't be the correct option either, as it doesn't match our equation. (D) \(x = 2 \tan y\): This option satisfies the equation. Thus, the correct answer is (D): \(x = 2 \tan y\).

Key concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals of products of functions. It is based on the product rule for differentiation and is formally stated by the rule \[ \int u dv = uv - \int v du \.\] In this method, the integrand is split into two parts: one part (\( u \)) is differentiated (yielding \( du \)), and the other part (\( dv \)) is integrated (yielding \( v \)). Then, these new expressions are substituted into the formula to achieve an easier integral. This technique is particularly useful when integrating products of algebraic, exponential, logarithmic and trigonometric functions.

In the context of differential equations for the JEE, recognizing when to apply integration by parts can simplify the process of finding solutions to certain types of problems. An example of its application would be in finding the integral of \( x \sin(2y) \), which could potentially be approached using integration by parts if it were necessary.

Initial Value Problem

An initial value problem involves a differential equation along with a condition (or set of conditions) that specifies the value of the unknown function at a particular point, termed as the initial condition. Solving an initial value problem means finding a function that satisfies the differential equation and the initial condition simultaneously.

For the exercise at hand, where a curve passes through \([1, (\pi / 4)]\), this point serves as the initial condition. As we integrate the differential equation to find the general solution, we include an arbitrary constant. Applying the initial condition, one can solve for this constant, yielding a unique solution --- the particular curve that not only follows the slope function defined by the differential equation but also passes through the specified point.

Separable Differential Equations

Separable differential equations are a class of differential equations that can be expressed as the product of two functions, each depending on a different variable. This form allows the equation to be separated by algebraic manipulation into two integrals: one involving only \( x \) and the other involving only \( y \). Essentially, you can write the differential equation in the form \[ \frac{dy}{dx} = g(y)h(x) \.\] The method to solve a separable differential equation is straightforward: rearrange the terms to isolate \( dy \) and \( dx \) on different sides, integrate both sides, and solve for \( y \) to find the general solution.

In the given exercise, by separating the variables and integrating, we are practicing this exact process to solve the differential equation. This method is particularly vital for students to master for the JEE, as separable differential equations are common in the calculus portion of the exam.

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions. These can often be more complex than standard integrals, especially when they involve products of different trigonometric functions or powers of trigonometric functions. There are various strategies to tackle these, such as using trigonometric identities to simplify the integrand, performing substitution, or using integration by parts.

In our particular problem, the integral of \( \tan(y) \) is encountered, which is a basic trigonometric integral solved by knowing or deriving the correct identity: \[ \int \tan(y) \: dy = \ln |\sec y| \.\] For students preparing for JEE, mastering such integrals is essential since trigonometric functions are ubiquitous in calculus problems. Emphasis should be placed on knowing key trigonometric integrals and appropriate techniques to solve them, which may include integration by parts or specific substitutions that simplify the integrals into a more manageable form.