Question

For which one of the following choices of \( N(x, y) \), is the equation \[ (e^x \sin y - 2y \sin x) \, dx + N(x, y) \, dy = 0 \] an exact differential equation?

• A. \( N(x, y) = e^x \sin y + 2 \cos x \)
• B. \( N(x, y) = e^x \cos y + 2 \cos x \)
• C. \( N(x, y) = e^x \cos y + 2 \sin x \)
• D. \( N(x, y) = e^x \sin y + 2 \sin x \)

Hint:

For an exact differential equation, check that the mixed partial derivatives of \( M(x, y) \) and \( N(x, y) \) match.

Answer 1

The Correct Option is B

Step 1: Verify the exactness condition.
The equation is exact if the mixed partial derivatives of the components of the equation are equal, i.e., \[ \frac{\partial}{\partial y}\left( M(x, y) \right) = \frac{\partial}{\partial x}\left( N(x, y) \right) \] where \( M(x, y) = e^x \sin y - 2y \sin x \) and \( N(x, y) \) is one of the provided options. Step 2: Compute the partial derivatives for each choice of \( N(x, y) \).
For \( M(x, y) = e^x \sin y - 2y \sin x \), \[ \frac{\partial M}{\partial y} = e^x \cos y - 2 \sin x \] Now, compute \( \frac{\partial N}{\partial x} \) for each \( N(x, y) \): 1. For option (A): \( N(x, y) = e^x \sin y + 2 \cos x \), \[ \frac{\partial N}{\partial x} = e^x \sin y - 2 \sin x \] 2. For option (B): \( N(x, y) = e^x \cos y + 2 \cos x \), \[ \frac{\partial N}{\partial x} = e^x \cos y - 2 \sin x \] 3. For option (C): \( N(x, y) = e^x \cos y + 2 \sin x \), \[ \frac{\partial N}{\partial x} = e^x \cos y + 2 \cos x \] 4. For option (D): \( N(x, y) = e^x \sin y + 2 \sin x \), \[ \frac{\partial N}{\partial x} = e^x \sin y + 2 \cos x \] Step 3: Check exactness.
We see that for option (B), \[ \frac{\partial M}{\partial y} = e^x \cos y - 2 \sin x \quad \text{and} \quad \frac{\partial N}{\partial x} = e^x \cos y - 2 \sin x \] match, confirming that option (B) satisfies the exactness condition. Final Answer: \[ \boxed{N(x, y) = e^x \cos y + 2 \cos x} \]