Question

The necessary and sufficient condition for the differential equation, $M(x, y) dx + N(x, y) dy = 0$ to be exact is given by

• A. $\frac{\partial M}{\partial x} = \frac{\partial N}{\partial y}$
• B. $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
• C. $\frac{\partial M}{\partial x} = \frac{\partial N}{\partial x}$
• D. $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial y}$

Hint:

To remember which variable to differentiate with respect to: differentiate $M$ (which is multiplied by $dx$) with respect to $y$, and differentiate $N$ (which is multiplied by $dy$) with respect to $x$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A first-order differential equation $M \, dx + N \, dy = 0$ is exact if its left-hand side is the total differential of some function $u(x, y)$, meaning $du = M \, dx + N \, dy$.

Step 2: Detailed Explanation:
If there exists a function $u(x, y)$ such that:
\[ du = M \, dx + N \, dy \]
Then, by the definition of total differential:
\[ M = \frac{\partial u}{\partial x} \quad \text{and} \quad N = \frac{\partial u}{\partial y} \]
According to Clairaut's Theorem on mixed partial derivatives, if the second-order partial derivatives are continuous:
\[ \frac{\partial^2 u}{\partial y \partial x} = \frac{\partial^2 u}{\partial x \partial y} \]
Substitute $M$ and $N$ into this relation:
\[ \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right) = \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right) \implies \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]
This condition is both necessary and sufficient for the differential equation to be exact.
Therefore, the correct condition is $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

Step 3: Final Answer
The correct option is (B).