Problem 23 - Advanced Engineering Mathematics

Question

Determine a region of the \(x y\) -plane for which the given differential equation would have a unique solution whose graph passes through a point \(\left(x_{0}, y_{0}\right)\) in the region. $$ \left(x^{2}+y^{2}\right) y^{\prime}=y^{2} $$

Step-by-Step Solution

Answer

The unique solution exists in the region where \((x, y) \neq (0, 0)\).

Step 1: Rewrite the Differential Equation in Standard Form

First, we rewrite the given differential equation \((x^2 + y^2)y' = y^2\) in the standard form for existence and uniqueness. Divide by \(x^2 + y^2\) on both sides to isolate \(y'\): \[y' = \frac{y^2}{x^2 + y^2}.\]

Step 2: Identify the Form of the Equation

The equation \(y' = \frac{y^2}{x^2 + y^2}\) is of the form \(y' = f(x, y)\) where \(f(x, y) = \frac{y^2}{x^2 + y^2}\). This is slightly simplified since it does not depend on \(x\).

Step 3: Determine Where \(f(x, y)\) and \(\partial f/\partial y\) Are Continuous

To check the conditions for a unique solution, the function \(f(x, y)\) and its partial derivative with respect to \(y\), \(\frac{\partial f}{\partial y}\), must be continuous around point \((x_0, y_0)\). Calculate \(\frac{\partial f}{\partial y}\): \[\frac{\partial f}{\partial y} = \frac{2y(x^2 + y^2) - y^2(2y)}{(x^2 + y^2)^2} = \frac{2xy^2}{(x^2 + y^2)^2}.\]Both \(f(x, y)\) and \(\frac{\partial f}{\partial y}\) are continuous as long as \(x^2 + y^2 eq 0\).

Step 4: Define the Region in the \(xy\)-Plane

The condition \(x^2 + y^2 eq 0\) implies that we must exclude the point \((0, 0)\) because this is where \(x^2 + y^2 = 0\), leading to an undefined function. Therefore, the region where \(f(x, y)\) is continuous and the differential equation has a unique solution is everywhere except at the origin. Thus, the region is the entire \(xy\)-plane except the origin: \[(x, y) eq (0, 0)\].

Key Concepts

Existence and Uniqueness Theorem

When tackling differential equations, the Existence and Uniqueness Theorem is a vital tool that ensures a solution can be found and that it is unique. This theorem provides conditions under which a differential equation will have a unique solution that passes through a given point

• The function \(f(x, y)\) in the equation must be continuous near the point of interest, \((x_0, y_0)\).
• The partial derivative of \(f\) with respect to \(y\), written as \(\frac{\partial f}{\partial y}\), also needs to be continuous around this point.

Both conditions are necessary to guarantee that a solution is not only possible but is uniquely determined by these initial conditions. This ensures that if you start at point \((x_0, y_0)\), there will be one and only one trajectory or path that the solution follows. In our example, we determined that these conditions apply everywhere except at the origin, where the function becomes undefined.

Continuous Functions

Continuity is a concept that plays a crucial role in understanding differential equations and their solutions. A function is continuous when small changes in the input result in small changes in the output, essentially ensuring there's no sudden jumps or breaks in the graph of the function.
In the context of our differential equation, the function \(f(x,y)\) that we derived in standard form is \(f(x,y) = \frac{y^2}{x^2 + y^2}\). Here, continuity means that \(f(x,y)\) smoothly responds to slight changes in the inputs \(x\) and \(y\), which prevents sudden, unpredictable behavior. In determining the regions of continuity for \(f\) and its partial derivative with respect to \(y\), we found that both are continuous as long as \(x^2 + y^2 eq 0\).
Thus, understanding where this continuity holds helps pinpoint where our differential equation will have a unique, continuous solution.

Partial Derivatives

Partial derivatives are powerful tools used to understand how changes in one variable affect a function, while keeping others constant. In differential equations, they provide insights into how sensitive a solution is to perturbations in initial conditions or the parameters of the equation.
For the given equation, we needed to find the partial derivative of the function \(f(x, y)\) with respect to \(y\). This is represented as \(\frac{\partial f}{\partial y}\) and reflects how \(f(x,y)\) changes as \(y\) changes, holding \(x\) constant:
\[\frac{\partial f}{\partial y} = \frac{2xy^2}{(x^2 + y^2)^2}\] This derivative must be continuous around the point of interest to satisfy the conditions of the Existence and Uniqueness Theorem. Hence, by ensuring that \(\frac{\partial f}{\partial y}\) does not become undefined, except at the origin, we confirm the critical region where the differential equation behaves predictably.