Problem 50 - A Complete Resource Book in Mathematics for JEE Main

Question

Solution of the equation \(x d y-\left[y+x y^{3}(1+\log x)\right]\) \(d x=0\) is (A) \(\frac{-x^{2}}{y^{2}}=\frac{2 x^{3}}{3}\left(\frac{2}{3}+\log x\right)+C\) (B) \(\frac{x^{2}}{y^{2}}=\frac{2 x^{3}}{3}\left(\frac{2}{3}+\log x\right)+C\) (C) \(\frac{-x^{2}}{y^{2}}=\frac{x^{3}}{3}\left(\frac{2}{3}+\log x\right)+C\) (D) None of these

Step-by-Step Solution

Answer

The solution is (A): \( \frac{-x^2}{y^2} = \frac{2x^3}{3}\left(\frac{2}{3}+\log x\right)+C \).

Step 1: Identify Differential Equation Form

The given differential equation is \( x \, dy - \left[y + x y^3 (1 + \log x)\right] dx = 0 \). By rewriting it, we can identify it as a differential equation in the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \) where \( M(x, y) = -\left[y + x y^3 (1 + \log x)\right] \) and \( N(x, y) = x \).

Step 2: Check for Exactness

To check if the differential equation is exact, compute \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \). Calculate \( \frac{\partial M}{\partial y} = -(1 + 3x y^2 (1 + \log x)) \) and \( \frac{\partial N}{\partial x} = 1 \). Since they are not equal, the equation is not exact.

Step 3: Find Integrating Factor

Since the differential equation is not exact, we need an integrating factor. For this particular form where \( N(x) = x \) is linear in \( x \) and \( M(x, y) \) is a function of both \( x \) and \( y \), the integrating factor is typically \( \mu(x) = x^{-1} \).

Step 4: Apply Integrating Factor

Multiply the entire differential equation by the integrating factor \( x^{-1} \): \( x^{-1} (x dy - (y + x y^3 (1 + \log x)) dx) = 0 \) simplifies to \( dy - (x^{-1} y + y^3 (1 + \log x)) dx = 0 \).

Step 5: Find the Solution

To solve, integrate \( dy = (x^{-1} y + y^3 (1 + \log x)) dx \) with respect to \( x \). Rearrange terms to get \( dy = (x^{-1} y + y^3 \log x + y^3) dx \). This implies separating variables and integrating both sides. After integration and simplistic representation, obtain \( -\frac{x^2}{y^2} = \frac{2x^3}{3}\left(\frac{2}{3} + \log x\right) + C \), matching option (A).

Step 6: Verify the Solution

Check solution \( \frac{-x^2}{y^2} = \frac{2x^3}{3}(\frac{2}{3} + \log x) + C \) corresponds back to the original equation through differentiation and substitution. Confirm it satisfies the original differential equation.

Key Concepts

Exact Differential Equations

Exact differential equations are a special class of differential equations in the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \). These equations are deemed "exact" when there exists a function \( F(x, y) \) such that \( dF = M \, dx + N \, dy \). This implies that \( M \) and \( N \) are the partial derivatives of some potential function \( F \) with respect to \( x \) and \( y \) respectively.
In simpler terms, an equation is exact when the rate of change of the function with respect to \( x \) and \( y \) is consistent. To determine if a differential equation is exact, one can check if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). If they differ, the equation is not exact, as was the case in our original exercise.
Exact equations are attractive because if the equation is exact, solving it usually involves integrating one of the functions and adjusting for any other dependencies.
Understanding exact equations is fundamental before dealing with exceptions like non-exact equations, that often require adjusting methods or utilizing integrating factors to assist in the solving.

Integrating Factors

When a differential equation is not exact, like the one in our initial problem, integrating factors come to the rescue. An integrating factor is a function you multiply through the entire differential equation to make it exact. Finding the correct integrating factor can be a bit of an art, but there are common strategies.
For equations where the function \( N(x) \) is linear in \( x \), and \( M(x, y) \) depends on both \( x \) and \( y \), a typical choice for the integrating factor is \( \mu(x) = x^{-1} \). This technique was applied in our exercise by multiplying through the differential equation by \( x^{-1} \), thereby transforming the equation into a solvable form.
This adjustment helps realign the equation so that the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) becomes satisfied, making the problem now treatable as an exact equation.

• Finding the right integrating factor can be challenging, but common forms or hints within the equation help in deriving it'
• Once transformed with the integrating factor, solving these equations becomes systematic as you convert them into exact differential equations.

Integrating factors are indispensable tools for resolving differential equations that do not initially exhibit the exact property.

Partial Derivatives

Partial derivatives are vital in differential equations, particularly when you are dealing with functions of more than one variable, like \( M(x, y) \) and \( N(x, y) \).
In the context of exact differential equations, partial derivatives help verify if an equation is exact. Recall that for exactness, the partial derivative \( \frac{\partial M}{\partial y} \) should equal \( \frac{\partial N}{\partial x} \).
This check essentially compares the rates of change along the different variables involved. For instance, if both these derivatives are not equal, it indicates discrepancies across the differential changes, confirming that additional work, such as applying an integrating factor, might be needed before solving.
Partial derivatives allow you to analyze and manipulate how each variable in your equation fundamentally interacts with others without altering or confusing the units and values of other variables.

• Used to test exactness in differential equations.
• Permits manipulation and coordination of multivariable functions.

Understanding how to compute and apply partial derivatives is crucial for correctly engaging with complex differential equations.