91Ó°ÊÓ

The Existence and Uniqueness Theorem for differential equations acts like a guide for determining when a differential equation has a unique solution. Suppose you have a differential equation in the form \( \frac{dy}{dx} = f(x, y) \). For this to have a unique solution through a point \( (x_0, y_0) \), two main conditions must be met:

• The function \( f(x, y) \) must be continuous around the point \( (x_0, y_0) \).
• The partial derivative \( \frac{\partial f}{\partial y} \) must also be continuous in that region.

When these criteria are satisfied, you can confidently say that there is a unique solution to the equation passing through your specified point. By ensuring continuity in both \( f \) and its partial derivative with respect to \( y \), we guarantee that small changes in initial conditions result in small changes in the solution, which is essential for uniqueness.

Partial derivatives help us understand how a function changes when its variables change. In the context of our differential equation \( \frac{dy}{dx} = y^{2/3} \), the partial derivative \( \frac{\partial f}{\partial y} \) tells us how \( f \), which is \( y^{2/3} \), changes with \( y \). To find this, differentiate \( y^{2/3} \) with respect to \( y \):

\[ \frac{\partial f}{\partial y} = \frac{2}{3} y^{-1/3} \]

This expression helps us determine the regions where uniqueness of the solutions might be compromised. Notice, for \( y = 0 \), this derivative becomes undefined because it involves division by zero. Consequently, the partial derivative exists and is continuous only when \( y eq 0 \). Recognizing where a partial derivative breaks continuity directly impacts our ability to determine unique solutions.

The concept of continuity is crucial in differential equations. A function is continuous if there are no abrupt changes or gaps. For the differential equation \( \frac{dy}{dx} = y^{2/3} \), \( f(y) = y^{2/3} \) is continuous for all real \( y \). This addresses half of our criteria for solution uniqueness.

However, we also need to look at the continuity of the partial derivative \( \frac{\partial f}{\partial y} = \frac{2}{3} y^{-1/3} \). Here, continuity breaks down at \( y = 0 \) due to the inverse relationship, as dividing by zero isn't allowed.
Thus, for the differential equation to hold a unique solution, it must be considered only in regions where \( y > 0 \) or \( y < 0 \). Essentially, whenever \( y eq 0 \), both \( y^{2/3} \) and its partial derivative are continuous, satisfying the uniqueness conditions laid out by the theorem.