Problem 60 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the equation \((2 x \log y) d x+\left(\frac{x^{2}}{y}+3 y^{2}\right) d y=0\) is (A) \(x^{2} \log y+y^{3}=c\) (B) \(y^{3} \log x+x^{3}=c\) (C) \(x^{2} \log y-y^{3}=c\) (D) None of these

Step-by-Step Solution

Answer

The solution is (A) \(x^2 \log y + y^3 = c\).

Step 1: Identify the Form of the Equation

The given equation \((2x \log y) dx + \left(\frac{x^2}{y} + 3y^2\right) dy = 0\) is a first-order differential equation. It can be rewritten as \(P(x, y) \, dx + Q(x, y) \, dy = 0\), where \(P = 2x \log y\) and \(Q = \frac{x^2}{y} + 3y^2\).

Step 2: Check for Exactness

To determine if this differential equation is exact, we check if the cross partial derivatives are equal: \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (2x \log y)\) and \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left(\frac{x^2}{y} + 3y^2\right)\). Calculate these to see if they are equal.

Step 3: Calculate Partial Derivatives

Calculate \(\frac{\partial P}{\partial y} = \frac{2x}{y}\) and \(\frac{\partial Q}{\partial x} = \frac{2x}{y}\). Since these partial derivatives are equal, the differential equation is exact.

Step 4: Find the Potential Function \( \Psi(x, y) \)

We find \( \Psi(x, y) \) such that \( \frac{\partial \Psi}{\partial x} = P(x, y) = 2x \log y\) and \( \frac{\partial \Psi}{\partial y} = Q(x, y) = \frac{x^2}{y} + 3y^2\).

Step 5: Integrate \(P\) with Respect to \(x\)

Integrate \(P(x, y) = 2x \log y\) with respect to \(x\):\[ \Psi(x, y) = \int 2x \log y \, dx = x^2 \log y + f(y) \]where \(f(y)\) is an arbitrary function of \(y\).

Step 6: Determine Function \(f(y)\)

Take the derivative of \(\Psi(x, y)\) found from Step 5 with respect to \(y\) and set it equal to \(Q(x, y)\):\[\frac{\partial}{\partial y}(x^2 \log y + f(y)) = \frac{x^2}{y} + f'(y)\]\[= \frac{x^2}{y} + 3y^2\].This implies \(f'(y) = 3y^2\). Integrate to find \(f(y) = y^3 + C\).

Step 7: Write the Final Solution

Substitute \(f(y)\) back into \(\Psi(x, y)\):\[ \Psi(x, y) = x^2 \log y + y^3 + C \].This implies the equation is \(x^2 \log y + y^3 = c\), where \(c\) is an arbitrary constant.

Step 8: Match with Given Options

We compare our solution \(x^2 \log y + y^3 = c\) with the provided options: (A) \(x^2 \log y + y^3 = c\), (B) \(y^3 \log x + x^3 = c\), (C) \(x^2 \log y - y^3 = c\), and (D) None of these. The solution matches most closely with option (A).

Key Concepts

First-order differential equations

First-order differential equations involve derivatives of one function dependent on a single variable. These equations are called 'first-order' because they include the first derivative, but do not involve higher-order derivatives. In general, a first-order equation can be expressed in the form:

• \( \frac{dy}{dx} = f(x, y) \)

where \( f(x, y) \) is a given function of \( x \) and \( y \).
First-order differential equations can often be solved using a variety of methods, such as separation of variables, integrating factors, or in this specific case, examining whether the equation is exact. Understanding the specific form and properties of the equation is crucial for determining the best solution strategy.

Exact differential equations

An exact differential equation arises when a given first-order differential equation can be expressed as a total differential of a function \( \Psi(x, y) \). This means there exists a function \( \Psi(x, y) \) such that:

• \( d\Psi = P(x, y) \, dx + Q(x, y) \, dy \)

where \( P \) and \( Q \) are continuously differentiable functions.
To verify if an equation is exact, we check whether the cross partial derivatives are equal, that is:

• \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \)

If this condition holds, then the equation is exact, and we can proceed to find the potential function \( \Psi(x, y) \). This involves integrating \( P \) over \( x \) and \( Q \) over \( y \) to piece together \( \Psi(x, y) \). The solution is typically expressed in the form \( \Psi(x, y) = c \), where \( c \) is a constant.

Integration techniques

Integration techniques are essential for solving exact differential equations. In this context, we integrate \( P(x, y) \) with respect to \( x \) and \( Q(x, y) \) with respect to \( y \) to construct the potential function \( \Psi(x, y) \). Here are a few important things to remember:

• When integrating \( P(x, y) \) with respect to \( x \), treat \( y \) as a constant and add an arbitrary function \( f(y) \) to account for any dependence on \( y \).
• Similarly, when integrating \( Q(x, y) \) with respect to \( y \), treat \( x \) as a constant.
• The derivative of the resulting function \( \Psi(x, y) \) concerning the other variable should match \( Q(x, y) \) or \( P(x, y) \) which confirms the function's correctness.

These steps help identify if the equation is properly handled, leading to a successful solution. Understanding and practicing these integration techniques improve problem-solving and comprehension of differential equations.