Problem 79 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Consider the differential equation \(y^{2} d x+\left(x-\frac{1}{y}\right) d y=0 .\) If \(y(1)=1\), then \(x\) is given by: (a) \(4-\frac{2}{y}-\frac{e^{\frac{1}{y}}}{e}\) (b) \(3-\frac{1}{y}+\frac{e^{\frac{1}{y}}}{e}\) (c) \(1+\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}\) (d) \(1-\frac{1}{y}+\frac{e^{\frac{1}{y}}}{e}\)

Step-by-Step Solution

Answer

Option (d) is correct: \(x = 1 - \frac{1}{y} + \frac{e^{1/y}}{e}\).

Step 1: Analyze the Differential Equation

The given differential equation is \(y^{2} dx + \left(x - \frac{1}{y}\right) dy = 0\). To solve the differential equation, our goal is to express it in a separable form or identify an exact form.

Step 2: Check for Exactness

Consider the equation to be of the form \(M(x,y) \, dx + N(x,y) \, dy = 0\), where \(M(x,y) = y^2\) and \(N(x,y) = x - \frac{1}{y}\). Compute the partial derivatives: \(\frac{\partial M}{\partial y} = 2y\) and \(\frac{\partial N}{\partial x} = 1\). Since \(\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x}\), the equation is not exact.

Step 3: Make the Equation Exact

To make the equation exact, find an integrating factor. For a function of \(y\), assume \(\mu(y) = y^k\). Multiply through by \(y^k\), then solve \(y^{k+2} dx + \left(y^k x - y^{k-1}\right) dy = 0\). Choose \(k = 2\) so the equation becomes \(y^4 dx + (y^2 x - y) dy = 0\) and check for exactness.

Step 4: Solve the Exact Differential Equation

With \(y^4 dx + (y^2 x - y) dy = 0\) exact, find a potential function \(F(x,y)\). Compute \(\frac{\partial F}{\partial x} = y^4\) yields \(F = y^4x + g(y)\). Also, \(\frac{\partial F}{\partial y} = y^2x - y = 4y^3x + g'(y)\); solve \(y^2x - y = 4y^3x\), giving \(g(y) = -e^{1/y}\). So, \(F(x, y) = y^4x - e^{1/y}\).

Step 5: Apply Initial Condition

Given \(y(1) = 1\), find the constant by plugging \((x, y) = (1, 1)\) into \(y^4x - e^{1/y} = C\). This results in \(1 - e = C\).

Step 6: Solve for x

Utilize the implicit solution \(y^4x - e^{1/y} = 1 - e\). Solve for \(x\) to get \(x = \frac{1 - e + e^{1/y}}{y^4}\). Simplify \(x = y - \frac{1}{y} + \frac{e^{1/y}}{e}\).

Step 7: Compare with Given Options

Compare the expression for \(x\) to the options provided. We find that \(x = y - \frac{1}{y} + \frac{e^{1/y}}{e}\) matches option (d).

Key Concepts

Exact Differential Equations

An exact differential equation is a type of differential equation where a single function can be identified whose differential equals the given equation. This means that the equation can be written as \( M(x,y) \, dx + N(x,y) \, dy = 0 \) and the function \( F(x,y) \) has the property that \( \frac{\partial F}{\partial x} = M(x,y) \) and \( \frac{\partial F}{\partial y} = N(x,y) \). To determine if a differential equation is exact, we check if the partial derivative of \( M \) with respect to \( y \), \( \frac{\partial M}{\partial y} \), equals the partial derivative of \( N \) with respect to \( x \), \( \frac{\partial N}{\partial x} \). If they are equal, the equation is exact, meaning it directly corresponds to a function on the plane. If they do not match, like in our case, the equation is not exact and further manipulation is needed, such as finding an integrating factor.

Integrating Factor

An integrating factor is a function, typically of \( x \) or \( y \) alone, which when multiplied with a non-exact differential equation, renders it exact. The idea is to adjust the original differential equation such that the derivatives match as needed for an exact equation. Often, we look for a function \( \mu(y) = y^k \) or similar. In our case, we chose \( \mu(y) = y^2 \). This affected the original factors in the equation, effectively transforming it into \( y^4 \, dx + (y^2 x - y) \, dy = 0 \). By doing so, the transformed equation now satisfies the exactness condition. The application of the integrating factor is a common method to handle non-exact differential equations, making them solvable in a straightforward manner.

Separable Differential Equations

Separable differential equations are a class of equations that can be arranged such that all terms involving one variable (and its differential) are on one side of the equation, and all terms involving the other variable are on the opposite side. This allows us to solve them by integration. The equation \( \frac{dy}{dx} = g(y)h(x) \) can be rearranged as \( \frac{1}{g(y)} \, dy = h(x) \, dx \), allowing both sides to be integrated separately. Although the original problem was not directly separable, achieving separability or an exact equation often relies on similar properties. Once separability is achieved or exactness corrected via an integrating factor, we proceed with integration as usual. Solving the resulting integrals provides us with a relation between the variables, which can often be simplified or solved given initial or specific boundary conditions.

Initial Value Problem

An initial value problem involves solving a differential equation subject to specific conditions at a given point, usually termed as an 'initial condition'. This specifies the value of the solution at a particular point, helping to find an exact solution rather than a general one. In our exercise, the initial condition given was \( y(1) = 1 \). By substituting this condition into the solution of the differential equation, we determine the constant of integration or adjust the solution to meet this criterion. This makes the solution unique to the situation described by the initial condition, transforming general solutions into precise, applicable real-world solutions. Hence, solving these problems involves finding solutions that satisfy both the equation and these initial criteria.