Question

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

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

Hint:

For a differential equation of the form \( M(x,y)dx + N(x,y)dy = 0 \) to be exact, the partial derivative of \(M\) with respect to \(y\) must be equal to the partial derivative of \(N\) with respect to \(x\). This is a fundamental condition for exactness, often remembered as "cross-differentiation" equality.

Answer 1

The Correct Option is C

Step 1: Understand the concept of an exact differential equation.
A first-order differential equation of the form \( M(x,y)dx + N(x,y)dy = 0 \) is said to be an exact differential equation if there exists a continuously differentiable function \( \phi(x,y) \) (also called a potential function) such that its total differential \( d\phi \) is equal to \( M(x,y)dx + N(x,y)dy \). The total differential of a function \( \phi(x,y) \) is given by: \[ d\phi = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy \] For the given equation \( M(x,y)dx + N(x,y)dy = 0 \) to be exact, we must have: \[ M(x,y) = \frac{\partial \phi}{\partial x} \quad \cdots (1) \] and \[ N(x,y) = \frac{\partial \phi}{\partial y} \quad \cdots (2) \]
Step 2: Derive the necessary condition.
For the function \( \phi(x,y) \) to exist, a fundamental property of continuous second partial derivatives (Clairaut's Theorem or Schwarz's Theorem) states that the order of differentiation does not matter, i.e., \[ \frac{\partial^2 \phi}{\partial y \partial x} = \frac{\partial^2 \phi}{\partial x \partial y} \] Now, we can substitute the expressions for \(M\) and \(N\) from equations (1) and (2) into this equality. Taking the partial derivative of Equation (1) with respect to \(y\): \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} \left( \frac{\partial \phi}{\partial x} \right) = \frac{\partial^2 \phi}{\partial y \partial x} \] Taking the partial derivative of Equation (2) with respect to \(x\): \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left( \frac{\partial \phi}{\partial y} \right) = \frac{\partial^2 \phi}{\partial x \partial y} \] For the equation to be exact, these mixed partial derivatives must be equal: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] This is the necessary (and sufficient, if \(M\) and \(N\) have continuous first partial derivatives) condition for a first-order differential equation to be exact.
Step 3: Compare with the given options.
The derived necessary condition for an exact differential equation is \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
Comparing this with the given options:
(1) \( \frac{\partial N}{\partial y} = \frac{\partial M}{\partial x} \) Incorrect. This is not the standard condition.
(2) \( \frac{\partial N}{\partial y} = - \frac{\partial M}{\partial x} \) Incorrect. This involves an incorrect sign and incorrect derivatives. (3) \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) Correct. This matches the derived condition.
(4) \( \frac{\partial M}{\partial y} = - \frac{\partial N}{\partial x} \) Incorrect. This involves an incorrect sign. The final answer is \( \boxed{\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}} \).