Question

If \( x^2 + xy^2 = c \), where \( c \in \mathbb{R} \), is the general solution of the exact differential equation \[ M(x, y)\, dx + 2xy\, dy = 0, \] then \( M(1,1) \) is ............

Hint:

To find \( M(x, y) \) in an exact differential equation, differentiate the given potential function and match coefficients with the differential form.

Answer 1

Correct Answer: 3

Step 1: Differentiate the given equation.
Given \( x^2 + xy^2 = c \), differentiating both sides: \[ 2x\,dx + (y^2 + 2xy\,dy) = 0. \] So, \[ (2x + y^2)\,dx + 2xy\,dy = 0. \]

Step 2: Compare with given form.
Given equation: \( M(x, y)\,dx + 2xy\,dy = 0. \) Thus, \( M(x, y) = 2x + y^2. \)

Step 3: Evaluate at (1,1).
\[ M(1,1) = 2(1) + (1)^2 = 3. \] However, as the equation was \( M\,dx + 2xy\,dy = 0 \), \( M \) is negative of what appears if rearranged to \( M\,dx = -2xy\,dy \), so effectively \( M(1,1) = -3. \)

Final Answer: \[ \boxed{-3} \]