Question

The initial value problem \[ \frac{dy}{dx} = \cos(xy), \quad x \in \mathbb{R}, \quad y(0) = y_0, \] where \( y_0 \) is a real constant, has

• A. a unique solution
• B. exactly two solutions
• C. infinitely many solutions
• D. no solution

Hint:

When solving initial value problems, always check the continuity of the function and its partial derivatives to ensure the existence of a unique solution using the existence and uniqueness theorem.

Answer 1

The Correct Option is A

This is a first-order ordinary differential equation of the form \( \frac{dy}{dx} = \cos(xy) \), with the initial condition \( y(0) = y_0 \). We apply the existence and uniqueness theorem to determine the nature of the solution. The theorem states that for an initial value problem of the form \( \frac{dy}{dx} = f(x, y) \) with an initial condition \( y(x_0) = y_0 \), if the function \( f(x, y) \) and its partial derivative with respect to \( y \) are continuous in a region containing \( (x_0, y_0) \), then a unique solution exists in some interval around \( x_0 \). Here, the function \( f(x, y) = \cos(xy) \) and its partial derivative with respect to \( y \) are both continuous for all values of \( x \) and \( y \). Specifically: \[ \frac{\partial}{\partial y} \cos(xy) = -x \sin(xy), \] which is continuous for all \( x \) and \( y \). Therefore, by the existence and uniqueness theorem, the initial value problem has a unique solution. Step 2: Final Answer.
The correct answer is (A) a unique solution.