Problem 81 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

Option (A) \(\frac{x}{y} - 2 \log x + 3 \log y = c\) is correct.

Step 1: Check if the Equation is Exact

Given the equation \((x^2 y - 2xy^2) \, dx - (x^3 - 3x^2 y) \, dy = 0\), we need to check if it is exact. An equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Here, \( M = x^2 y - 2xy^2 \) and \( N = -(x^3 - 3x^2 y) \). Compute the partial derivatives: \( \frac{\partial M}{\partial y} = x^2 - 4xy \) and \( \frac{\partial N}{\partial x} = -3x^2 + 6xy \). Since these are not equal, the equation is not exact.

Step 2: Check for an Integrating Factor

Since the differential equation is not exact, we look for an integrating factor that makes it exact. Calculate \( \frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) \). Substituting \( M = x^2 y - 2xy^2 \) gives us an integrating factor that depends on \( x \), \( \mu(x) = x^{-2} \). Multiplying the entire differential equation by this integrating factor should make it exact.

Step 3: Verify the Equation with the Integrating Factor

Multiply the original differential equation by the integrating factor \( x^{-2} \): \( (y - 2y^2) \, dx - (x - 3y) \, dy = 0 \). Now, check exactness: provide that this new equation satisfies \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). It turns out exact with the integrating factor, allowing solution finding.

Step 4: Solve the Exact Differential Equation

After confirming exactness, solve by integrating \( M = y - 2y^2 \) with respect to \( x \) and \( N = -(x - 3y) \) with respect to \( y \). The solution must total: \( \frac{x}{y} - 2\log(x) + 3\log(y) = C \). Match this to the original problem choices.

Step 5: Check Against Solution Options

Compare the resulting expression to the given solution options: (A) \(\frac{x}{y} - 2 \log x + 3 \log y = c\). Our derived expression matches option (A) entirely, confirming it as the correct solution.

Key Concepts

Exact Equations

An exact differential equation has a special structure where the solution can be found more straightforwardly. An equation is said to be exact if the differential expression, say \(M\,dx + N\,dy = 0\), satisfies the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). This condition ensures that there is a potential function \(F(x,y)\) such that its total differential gives back the original equation:

• The function \(M = M(x, y)\) is the partial derivative of \(F\) with respect to \(x\).
• The function \(N = N(x, y)\) is the partial derivative of \(F\) with respect to \(y\).

If these partial derivatives are equal, the differential equation is exact, meaning \(F(x, y) = C\) (where \(C\) is a constant) is the solution to the differential equation.
When an equation is not exact, an integrating factor might be used to make it exact, such as multiplication by a certain function that equalizes \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\). This facilitates finding the potential function.

Integrating Factor

When a differential equation is not exact, you can sometimes transform it into an exact equation using an integrating factor. An integrating factor is a function that, when multiplied by the entire equation, creates equal partial derivatives, thereby making it exact.
Here's how you can identify and use an integrating factor:

• First, calculate \(\frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\). If this expression depends only on \(x\), the integrating factor will also only depend on \(x\).
• Determine the integrating factor \(\mu(x)\) by solving the derived expression. In the solution provided, \(\mu(x) = x^{-2}\) was used.
• Multiply the entire original equation by this integrating factor. This transforms the differential equation into an exact form.

After transforming the equation, you can proceed to solve it for the function \(F(x, y) = C\), which reflects the solution of the original problem.

Partial Derivatives

Partial derivatives are a fundamental tool in dealing with functions of multiple variables, such as in differential equations. They measure how a function changes as one of the input variables changes while holding others constant.
In an exact equation, you need to compute these derivatives to check whether the equation is exact:

• If \(M = M(x,y) = x^2 y - 2xy^2\), then \(\frac{\partial M}{\partial y} = x^2 - 4xy\) shows how \(M\) changes with respect to \(y\).
• For \(N = N(x,y) = -(x^3 - 3x^2 y)\), \(\frac{\partial N}{\partial x} = -3x^2 + 6xy\) is computed to check exactness.

Only when these partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) are equal or can be made equal through an integrating factor does the differential equation become solvable with a straightforward method, leading to the solution \(F(x, y) = C\). Partial derivatives help navigate and manipulate equations so that solutions can be effectively determined.