Problem 26 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(y=y(x)\) be a solution of the differential equation, \(\sqrt{1-x^{2}} \frac{d y}{d x}+\sqrt{1-y^{2}}=0, \mid x k 1\) If \(y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}\), then \(y\left(\frac{-1}{\sqrt{2}}\right)\) is equal to: (a) \(\frac{\sqrt{3}}{2}\) (b) \(-\frac{1}{\sqrt{2}}\) (c) \(\frac{1}{\sqrt{2}}\) (d) \(-\frac{\sqrt{3}}{2}\)

Step-by-Step Solution

Answer

The answer is (c) \(\frac{1}{\sqrt{2}}\).

Step 1: Rearrange the Differential Equation

Start by rearranging the given differential equation:\[ \sqrt{1-x^{2}} \frac{dy}{dx} + \sqrt{1-y^{2}} = 0 \]Rewriting it gives:\[ \sqrt{1-x^2} \frac{dy}{dx} = -\sqrt{1-y^2} \]Now, separate variables for integration by dividing both sides by \(\sqrt{1-y^2}\):\[ \frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}} \]

Step 2: Integrate Both Sides

Now, integrate both sides of the equation:\[ \int \frac{dy}{\sqrt{1-y^2}} = \int -\frac{dx}{\sqrt{1-x^2}} \]The integral of \(\frac{1}{\sqrt{1-u^2}}\) with respect to \(u\) is \(\arcsin(u)\). Thus:\[ \arcsin(y) = -\arcsin(x) + C \]

Step 3: Apply the Initial Condition

Use the initial condition given, \(y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}\), to find \(C\):\[ \arcsin\left(\frac{\sqrt{3}}{2}\right) = -\arcsin\left(\frac{1}{2}\right) + C \]\[ \frac{\pi}{3} = -\frac{\pi}{6} + C \]Solving for \(C\) gives:\[ C = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2} \]

Step 4: Solve for the Desired Value

Now use the solved equation \(\arcsin(y) = -\arcsin(x) + \frac{\pi}{2}\) and plug in \(x = -\frac{1}{\sqrt{2}}\):\[ \arcsin(y) = -\arcsin\left(-\frac{1}{\sqrt{2}}\right) + \frac{\pi}{2} \]Recall \(\arcsin\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4} \), so:\[ \arcsin(y) = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \]This implies \(y = \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}\).

Key Concepts

Separation of Variables

Separation of variables is a vital technique used to solve differential equations where variables can be rearranged so that each one appears on a different side of the equation. Imagine it like untangling a knot, dividing it into smaller, manageable sections which can be handled more easily. In the exercise, our differential equation \[ \sqrt{1-x^{2}} \frac{dy}{dx} + \sqrt{1-y^{2}} = 0\]is initially rearranged to group all terms involving \(x\) on one side and \(y\) on the other:\[ \frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}} \]Through this process, the complex interconnected relationship between \(x\) and \(y\) in a differential equation is split. This enables an easier path to integration on both sides, helping to ultimately find a functional solution. Whenever the variables can be separated so cleanly, it allows for neat, straightforward integrations, a blessing in mathematics!

Initial Value Problem

An initial value problem is a differential equation accompanied by a specific value, called an initial condition. This extra piece of information helps us determine a unique solution from a family of possible solutions. In the given problem, after separating the variables and integrating, we find a general solution: \[\arcsin(y) = -\arcsin(x) + C\]Here, \(C\) is a constant, representing an infinite range of possible curves that satisfy the equation. To anchor our solution to a specific scenario, we employ the initial condition provided: \(y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}\). Substituting these values allows us to solve for \(C\):

• Substitute \( \arcsin\left(\frac{\sqrt{3}}{2}\right) = -\arcsin\left(\frac{1}{2}\right) + C \)
• This becomes \( \frac{\pi}{3} = -\frac{\pi}{6} + C \)
• Solving gives \( C = \frac{\pi}{2} \)

This step ensures we have a specific, single path our solution follows. Initial value problems like this are crucial because they often model real-world phenomena, where initial states provide necessary insight into system behavior.

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions, often arising when solving differential equations involving circular functions like sine and cosine. In this problem, after separating the variables, we integrate both sides:\[\int \frac{dy}{\sqrt{1-y^2}} = \int -\frac{dx}{\sqrt{1-x^2}}\]The integration of \( \frac{1}{\sqrt{1-u^2}} \) gives us \( \arcsin(u) \). Similarly, integrating both sides results in:

• \( \arcsin(y) \) on one side.
• \( -\arcsin(x) + C \) on the other side.

These integrals require familiarity with trigonometric identities and their derivatives. Integration here utilizes the fact that sine and cosine functions are derivatives of each other. Such integrals are common in physics and engineering, modeling wave forms and oscillatory behavior. Mastering them can elegantly solve many kinds of practical problems!