Problem 19 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The solution curve of the differential equation, \(\left(1+e^{-x}\right)\left(1+y^{2}\right) \frac{d y}{d x}=y^{2}\), which passes through the point \((0,1)\), is: (a) \(y^{2}+1=y\left(\log _{e}\left(\frac{1+e^{-x}}{2}\right)+2\right)\) (b) \(y^{2}+1=y\left(\log _{e}\left(\frac{1+e^{x}}{2}\right)+2\right)\) (c) \(y^{2}=1+y \log _{e}\left(\frac{1+e^{x}}{2}\right)\) (d) \(y^{2}=1+y \log _{e}\left(\frac{1+e^{-x}}{2}\right)\)

Step-by-Step Solution

Answer

The correct answer is option (d).

Step 1: Identify the Differential Equation

The given differential equation is \((1 + e^{-x})(1 + y^2) \frac{dy}{dx} = y^2\). We also know that the solution curve passes through the point \((0, 1)\).

Step 2: Separate Variables

To solve the equation, separate the variables by rewriting it as: \( (1 + y^2) \frac{dy}{y^2} = \frac{1}{1 + e^{-x}} dx \). This separates the differentials such that one side involves only \(y\) and \(dy\), and the other side involves only \(x\) and \(dx\).

Step 3: Integrate Both Sides

Integrate both sides: \( \int \frac{1 + y^2}{y^2} \, dy = \int \frac{1}{1 + e^{-x}} \, dx \). This gives two integrals: \( \int \left( \frac{1}{y^2} + 1 \right) \, dy \) on the left and \( \int \frac{1}{1 + e^{-x}} \, dx \) on the right.

Step 4: Solve the Left Integral

The left integral simplifies to \( \int \left( \frac{1}{y^2} + 1 \right) \, dy = \int y^{-2} \, dy + \int 1 \, dy \). Computing these, we get \(-y^{-1} + y = -\frac{1}{y} + y\).

Step 5: Solve the Right Integral

The right integral, \( \int \frac{1}{1 + e^{-x}} \, dx \), requires a substitution. Let \( u = 1 + e^{-x} \), then \( du = -e^{-x} \, dx \) or \( dx = -\frac{du}{e^{-x}} = -e^x \, du \).Substitute and integrate: \( \int \frac{1}{u} \, (-e^x \, du) = -\int \frac{e^x}{u} \, du = -\int \, du = -\log|u| + C \). Replace \(u\) and simplify: \(-\log|1 + e^{-x}| + C\).

Step 6: Combine Solution

Combining both sides, we have: \(-\frac{1}{y} + y = -\log|1 + e^{-x}| + C\).

Step 7: Apply the Initial Condition

Use the condition \((0,1)\) to find \(C\). Substitute \(x = 0\) and \(y = 1\) into the equation: \[-1 + 1 = -\log|1 + e^{0}| + C \quad \Rightarrow \quad 0 = -\log(2) + C \]. Thus, \(C = \log(2)\).

Step 8: Write the Final Solution

Substitute \(C\) back into the solution: \[-\frac{1}{y} + y = -\log|1 + e^{-x}| + \log(2) \] which implies \[ y + \frac{1}{y} = \log \left(\frac{1 + e^{-x}}{2} \right) \]. Rearranging gives \[ y^2 = 1 + y \log \left(\frac{1 + e^{-x}}{2} \right) \].

Key Concepts

Separation of Variables

Separation of variables is a fundamental technique used in solving differential equations. In this problem, our main goal is to manipulate the differential equation in such a way that both sides contain only one variable each. This makes it possible to integrate both sides easily.

The given differential equation is \((1 + e^{-x})(1 + y^2) \frac{dy}{dx} = y^2\). To solve it, we need to "separate the variables." Essentially, we rearrange the equation to isolate terms involving \(y\) on one side and terms involving \(x\) on the other side. This process involves some algebra and insight into how the variables interact.

• First, divide both sides by \((1 + y^2)\) and \(y^2\) to isolate \(dy\) and \(dx\):
  \(\frac{(1 + y^2)\, dy}{y^2} = \frac{dx}{1 + e^{-x}}\).
• Now, the variables are separated: \(dy\) and \(y\) terms are on the left, while \(dx\) and \(x\) terms are on the right.

By achieving this separation, we can directly integrate with respect to \(y\) and \(x\) respectively, a crucial step towards solving for the function \(y(x)\).

Integration Techniques

Once we've separated the variables in the differential equation, our next task is to integrate both sides. Each side of the equation offers its own set of challenges and requires specific techniques to integrate effectively.

For the left side of the equation, we are integrating \( \int \left( \frac{1}{y^2} + 1 \right) \, dy \). Breaking it down:

• The term \( \int y^{-2} \, dy \) simplifies into \(-y^{-1}\), because the antiderivative of \(a^{-n}\) is \(-a^{-n+1}/(n-1)\).
• The simple linear term \( \int 1 \, dy \) gives \(y\), as integrating a constant simply appends that constant with the variable.

The integration of the right side \( \int \frac{1}{1 + e^{-x}} \, dx \) is more complex and involves substitution:

• Use substitution \( u = 1 + e^{-x} \), leading to \( du = -e^{-x} \, dx \).
• Replace \(dx\) with \(-e^x \, du\) to simplify integration into \(-\int \frac{1}{u} \, du\).
• This yields \(-\log|u| + C\). Replace \(u\) back with \(1 + e^{-x}\) post-integration.

These integration steps are critical for solving the differential equation and forming a clearer picture of \(y\) as a function of \(x\).

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation problem that provides specific conditions at an initial point. Typically, these constraints are needed to solve for a particular solution among a family of potential solutions.

In this exercise, we are given the condition \((0, 1)\). This means that when \(x = 0\), \(y\) is expected to be 1. Such a condition helps determine the constant of integration, making the solution applicable to the specific problem context.

• Once we've formed the general solution from integration, say \(-\frac{1}{y} + y = -\log|1 + e^{-x}| + C\), we substitute \(x = 0\) and \(y = 1\).
• This points us towards solving \(0 = -\log(2) + C\), leading to \(C = \log(2)\).

Substituting back this constant into the solution allows us to form a particular solution \(y^2 = 1 + y \log \left(\frac{1 + e^{-x}}{2} \right)\), which fits the original problem's specific initial condition. Recognizing and applying initial conditions accurately helps ensure the solution aligns exactly with the given data points.