Problem 80 - A Complete Resource Book in Mathematics for JEE Main

Question

A differential equation is said to be exact if it can be derived from its primitive by direct differentiation without any further transformation such as elimination, etc. The differential equation $$ \left(x^{2}-a y\right) d x+\left(y^{2}-a x\right) d y=0 $$ is exact in as much as it can be derived from its primitive $$ x^{3}-3 a x y+y^{3}=c $$ by direct differentiation. The necessary and sufficient condition for the differential equation \(M d x+N d y=0\) to be exact is that $$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $$ where \(\frac{\partial M}{\partial y}\) is the derivative of \(M\) with respect to \(y\) keeping \(x\) as constant and \(\frac{\partial N}{\partial x}\) is the derivative of \(N\) with respect to \(x\) keeping \(y\) as constant. If the equation \(M d x+N d y=0\) is exact, then it can be integrated as follows: Firstly, integrate \(M\) with respect to \(x\) regarding \(y\) as constant. Then, integrate with respect to \(y\) those of the terms in \(N\) which do not involve \(x\). The sum of the two expressions thus obtained equated to a constant is the required solution. The solution of the equation \(\left[y\left(1+\frac{1}{x}\right)+\cos y\right] d x+[x+\log x-x \sin y] d y=0 i s\) (A) \(y(x+\log x)-x \cos y=c\) (B) \(y(x+\log x)+x \sin y=c\) (C) \(y(x+\log x)+x \cos y=c\) (D) None of these

Step-by-Step Solution

Answer

The solution to the differential equation is (C) \(y(x+\log x) + x \cos y = c\).

Step 1: Differentiate Each Part

To determine if the given differential equation is exact, calculate \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\). Here, \(M = y\left(1 + \frac{1}{x}\right) + \cos y\) and \(N = x + \log x - x \sin y\). Compute- \(\frac{\partial M}{\partial y} = \frac{d}{dy}\left[y\left(1+\frac{1}{x}\right) + \cos y\right] = \left(1+\frac{1}{x}\right) - \sin y\)- \(\frac{\partial N}{\partial x} = \frac{d}{dx}[x + \log x - x \sin y] = 1 + \frac{1}{x} - \sin y\).

Step 2: Verify Exactness

Verify if the equation is exact by checking if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Both partial derivatives simplify to \(1 + \frac{1}{x} - \sin y\), confirming the differential equation is exact.

Step 3: Integrate M with Respect to x

Integrate \(M\) with respect to \(x\) while treating \(y\) as constant. Compute \( \int \left[y\left(1+\frac{1}{x}\right) + \cos y\right] dx = x y + y \log x + x \cos y + f(y)\), where \(f(y)\) is a function of \(y\).

Step 4: Integrate Terms in N Not Containing x with Respect to y

Identify terms in \(N\) that do not include \(x\) (after considering \(M\)): none are left, as they are embedded with \(y\) which are already integrated above. Thus, no separate integral is needed from \(N\).

Step 5: Form the General Solution

Combine the results from Steps 3 and 4. The integrated terms involve all possible variables, which sums up as:\[ y(x + \log x) + x \cos y = c \]

Step 6: Match with Given Options

Compare the derived solution to the provided options. The solution matches option (C): \(y(x+\log x)+x \cos y=c\).

Key Concepts

Partial Derivative

In calculus, partial derivatives play a key role in dealing with functions of multiple variables. They measure how a function changes as each of its input variables changes while keeping the other variables constant. For a function \(f(x, y)\), the partial derivative with respect to \(x\), denoted as \(\frac{\partial f}{\partial x}\), gathers information about the rate of change of \(f\) in the direction of the \(x\)-axis. Similarly, \(\frac{\partial f}{\partial y}\) does the same for the \(y\)-axis.

When analyzing differential equations, especially exact ones, it's crucial to compute these partial derivatives to determine the equation's exactness.

• This involves ensuring that \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), where \(M\) and \(N\) are functions in the differential equation \(Mdx + Ndy = 0\).
• This condition is both necessary and sufficient for the equation to be classified as exact.

In our specific example, partial derivatives confirm the equation's exact nature by showing equal derivatives, \(1 + \frac{1}{x} - \sin y\), emphasizing the importance of precision in this computation to solve exact differential equations effectively.

Differential Equation Solution

Finding the solution to a differential equation involves integrating and computing the relationships between variables within the equation. An exact differential equation simplifies this process by allowing direct integration due to its precise symmetrical properties between its partial derivatives.

• For an equation \(Mdx + Ndy = 0\), achieving exactness (i.e., \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)) allows us to proceed with integration strategies without further transformation.
• Solutions involve integrating each component with respect to its respective variable, aligning fundamentally with the symmetry noticed in exactness checks.

The step-by-step solution entails first integrating \(M\) with respect to \(x\), treating \(y\) as a constant. In this problem, this yields \(x y + y \log x + x \cos y\). Next, we examine \(N\) for any terms that are independent of \(x\), ensuring all paths are covered. Upon solving, we match our integrated results against provided choices, confirming solutions logically via exactness symmetry.

Integration in Differential Equations

The art of integration in differential equations is about skillfully combining functions and constants to reveal the hidden relationships within the equation. The goal is to bring out clear and simplified expressions that hold true over a range of conditions. For exact differential equations, the integration process is streamlined by leveraging the inherent balance in the problem.

The approach:

• Firstly, recognize that integration involves accumulation, adding up small parts to find the whole, just like piecing together a puzzle.
• Begin by integrating \(M\) fully with respect to \(x\), as shown in our example, which treats \(y\) as a constant. The result -- \(x y + y \log x + x \cos y\) -- covers the full range of functions contributed by \(M\).

Check whether all parts are accounted for by also evaluating \(N\) for isolated \(y\)-dependent terms, ensuring no dimension is neglected. This robustly discloses the sought-after solution, tying in constants to form the general equation:
\[ y(x + \log x) + x \cos y = c \]
Each integration step elegantly layers the final expression, building toward the ultimate solution.