Question

If \( M, N, \mu, w : \mathbb{R}^2 \to \mathbb{R} \) are differentiable functions with continuous partial derivatives, satisfying \[ \mu(x, y) M(x, y) \, dx + \mu(x, y) N(x, y) \, dy = dw, \] then which one of the following is TRUE?

• A. \( \mu w \) is an integrating factor for \( M(x, y) \, dx + N(x, y) \, dy = 0 \)
• B. \( \mu w^2 \) is an integrating factor for \( M(x, y) \, dx + N(x, y) \, dy = 0 \)
• C. \( w(x, y) = w(0,0) + \int_0^x \mu M(s) \, ds + \int_0^y \mu N(t) \, dt \), for all \( (x, y) \in \mathbb{R}^2 \)
• D. \( w(x, y) = w(0,0) + \int_0^x \mu M(s, y) \, ds + \int_0^y \mu N(x, t) \, dt \), for all \( (x, y) \in \mathbb{R}^2 \)

Hint:

When solving for integrating factors, look for terms that multiply the differential equation to simplify it into an exact form.

Answer 1

The Correct Option is A

Step 1: Analyze the given equation.
The given equation is: \[ \mu(x, y) M(x, y) \, dx + \mu(x, y) N(x, y) \, dy = dw. \] This is a general form of a first-order linear differential equation, and we are looking for the integrating factor. Step 2: Integrating Factor.
For a differential equation of the form: \[ M(x, y) \, dx + N(x, y) \, dy = 0, \] the solution can often be simplified by multiplying through by an integrating factor \( \mu(x, y) \). The given equation suggests that \( \mu(x, y) \) is indeed an integrating factor for this equation. Step 3: Conclusion.
The correct answer is \( \mu w \) as the integrating factor, as it satisfies the equation. Final Answer: \[ \boxed{\mu w \text{ is an integrating factor for } M(x, y) \, dx + N(x, y) \, dy = 0.} \]