Problem 21 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(y=y(x)\) be the solution of the differential equation, \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1 .\) If \(y(\pi)=a\) and \(\frac{d y}{d x}\) at \(x=\pi\) is \(b\), then the ordered pair \((a, b)\) is equal to : (a) \(\left(2, \frac{3}{2}\right)\) (b) \((1,-1)\) (c) \((1,1)\) (d) \((2,1)\)

Step-by-Step Solution

Answer

The ordered pair \((a, b) = (1, 1)\), option (c).

Step 1: Separate Variables

The given differential equation is \( \frac{2 + \sin x}{y+1} \cdot \frac{d y}{d x} = -\cos x \). First, separate the variables by multiplying both sides by \( (y + 1) \) and dividing by \(-\cos x\) to get: \( \frac{d y}{y + 1} = -\frac{\cos x}{2 + \sin x} \, dx \).

Step 2: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{d y}{y + 1} = \int -\frac{\cos x}{2 + \sin x} \, dx \). The left side integrates to \( \ln |y + 1| \). For the right side, use substitution \( u = 2 + \sin x \), giving \( du = \cos x \, dx \), leading to \( \int -\frac{1}{u} \, du \). This integrates to \( -\ln |u| = -\ln |2 + \sin x| \).

Step 3: Simplify the Integral Result

Result from integration: \( \ln |y + 1| = -\ln |2 + \sin x| + C \). Simplify: \( |y + 1| = \frac{C}{|2 + \sin x|} \). Since \( y > 0 \), \( y + 1 \) is positive, and we can drop the absolute values: \( y + 1 = \frac{C}{2 + \sin x} \).

Step 4: Use Initial Condition to Find C

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

Step 5: Solve for y(x) using C

Plug \( C = 4 \) back into the equation: \( y + 1 = \frac{4}{2 + \sin x} \). Thus \( y(x) = \frac{4}{2 + \sin x} - 1 \).

Step 6: Calculate y(\pi) to find a

Substitute \( x = \pi \) into \( y(x) \): \( y(\pi) = \frac{4}{2 + \sin \pi} - 1 = \frac{4}{2 + 0} - 1 = 2 - 1 = 1 \). So \( a = 1 \).

Step 7: Differentiate y(x) to find dy/dx

Differentiate \( y(x) = \frac{4}{2 + \sin x} - 1 \) with respect to \( x \) using the quotient rule: \( \frac{d}{dx} \left( \frac{4}{2 + \sin x} \right) = -4 \cdot \frac{\cos x}{(2 + \sin x)^2} \). Hence, \( \frac{dy}{dx} = -4 \cdot \frac{\cos x}{(2 + \sin x)^2} \).

Step 8: Calculate \( \frac{dy}{dx} \) at x=\pi to find b

Substitute \( x = \pi \) into \( \frac{dy}{dx} \): \( \frac{dy}{dx} \Big|_{x = \pi} = -4 \cdot \frac{\cos \pi}{(2 + \sin \pi)^2} = -4 \cdot \frac{-1}{4} = 1 \). So, \( b = 1 \).

Step 9: Form the Ordered Pair \((a,b)\)

Combine results to form the ordered pair \((a, b)\): \( (1, 1) \). This corresponds to option (c).

Key Concepts

Initial Value Problem

In the world of differential equations, an initial value problem (IVP) is a type of problem where you're tasked with finding a function that fulfills two criteria:

• It satisfies a given differential equation.
• It also meets a specific initial condition.

Think of it like this: you're given a road map (the differential equation) and a starting point (the initial condition), and your job is to figure out the path that begins at the starting point and strictly follows the map.
For example, in our given exercise, the differential equation is \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x\), and the initial condition provided is \(y(0) = 1\). This means that at the start, when \(x = 0\), the function's value is \(1\).
Solving an IVP typically involves finding a general solution to the differential equation first, and then using the initial condition to ascertain any constants involved. This ensures the solution is particular to the specific situation you're studying.

Separation of Variables

Separation of variables is a key technique for solving differential equations, particularly when they can be split into two separate parts: one involving only the dependent variable, and the other only the independent variable. It’s a powerful tool because it turns a daunting differential equation into something much simpler to handle.
In our specific problem, we start with the equation \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x\). To employ separation of variables, we rearrange terms to isolate like variables on either side, resulting in \(\frac{d y}{y + 1} = -\frac{\cos x}{2 + \sin x} \, dx\). This new equation presents a form where you can integrate each side with respect to its own variable.
This separation transforms the original equation into two simpler integrals, making it easier to solve and analyze each variable individually.

Integration Techniques

Integration is an integral part (pun intended!) of solving differential equations. Once you separate the variables, integration techniques allow you to derive the function that fits both the differential equation and the initial condition.
With our separated equation \(\int \frac{d y}{y + 1} = \int -\frac{\cos x}{2 + \sin x} \, dx\), the left-hand side integrates straightforwardly to \(\ln |y + 1|\). The right-hand side requires a bit more ingenuity, where we use a substitution technique.

• By letting \(u = 2 + \sin x\), we calculate \(du = \cos x \, dx\), turning the integral into a simpler one \(\int -\frac{1}{u} \, du\).
• This integrates to \(-\ln |u| = -\ln |2 + \sin x|\).

After integrating, we piece together these parts back into the form \(\ln |y + 1| = -\ln |2 + \sin x| + C\) and solve for the particular solution using the initial condition. Integration not only provides solutions, but also highlights the elegance and depth of calculus applied in differential equations.