Problem 24 - 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. $$ (y-x) y^{\prime}=y+x $$

Step-by-Step Solution

Answer

Unique solutions exist in the region \(y \neq x\).

Step 1: Identify the Differential Equation

The given differential equation is \((y-x) y' = y + x\). The equation is in the form of \((y-x)\frac{dy}{dx} = y + x\), where \(y' = \frac{dy}{dx}\). This is a first-order differential equation.

Step 2: Rearrange the Differential Equation

Rearrange the equation to express \(y'\) in terms of \(x\) and \(y\):Divide both sides by \((y-x)\) (assuming \(y eq x\)), \[\frac{dy}{dx} = \frac{y+x}{y-x}\].

Step 3: Check Conditions for Unique Solutions (Existence and Uniqueness Theorem)

The existence and uniqueness theorem requires that the functions \(f(x, y)\) and its partial derivative \(\frac{\partial f}{\partial y}\) be continuous in some region around \((x_0, y_0)\). Here, \(f(x, y) = \frac{y+x}{y-x}\).

Step 4: Calculate the Partial Derivative

Find \(\frac{\partial f}{\partial y}\): \[f(x, y) = \frac{y+x}{y-x} = \frac{1 + \frac{x}{y}}{1 - \frac{x}{y}}\] simplifying gives,\[\frac{\partial f}{\partial y} = -\frac{2x}{(y-x)^2}\]. This derivative exists and is continuous everywhere except at \(y=x\), where it is undefined.

Step 5: Determine the Region of the \(xy\)-Plane

The function \(f(x, y)\) is not continuous along the line \(y = x\), so the region where a unique solution exists must exclude \(y = x\). Therefore, the region is defined as all points \(\{(x,y) \mid y eq x\}\).

Step 6: Conclusion on Region's Properties

For any point \((x_0, y_0)\) that lies in the region \(y eq x\), the differential equation will have a unique solution passing through that point.

Key Concepts

Existence and Uniqueness Theorem

In differential equations, the Existence and Uniqueness Theorem is an important concept that helps us determine if a differential equation has a unique solution passing through a specific point. This theorem tells us that for a first-order differential equation \(dy/dx = f(x, y)\), there exists a unique solution if two conditions are satisfied.

First, the function \(f(x, y)\) must be continuous in a certain region around the point \((x_0, y_0)\). Second, the partial derivative of \(f(x, y)\) with respect to \(y\), written as \(\partial f/\partial y\), must also be continuous in this region. If both these conditions hold, any initial value problem will have exactly one solution that passes through the initial point \((x_0, y_0)\).

This theorem helps us avoid ambiguity in solving differential equations, ensuring we find a unique path according to the specified conditions.

Partial Derivative

The concept of a partial derivative is fundamental in calculus, especially when dealing with functions of multiple variables. In this context, we look at how a function changes as we vary one of its variables while keeping others constant.

For the differential equation \((y-x) y' = y + x\), we derived the function \(f(x, y) = \frac{y + x}{y - x}\). To apply the Existence and Uniqueness Theorem, we needed the partial derivative of \(f(x, y)\) with respect to \(y\), noted as \(\frac{\partial f}{\partial y}\).

By simplifying and differentiating, we found \(\frac{\partial f}{\partial y} = -\frac{2x}{(y-x)^2}\). This function is continuous wherever \(y eq x\). Understanding partial derivatives is crucial because they reveal how the solution changes with small variations in one variable while keeping the other fixed.

Continuous Function

A continuous function is one that doesn't have any abrupt jumps, breaks, or holes. In the context of the Existence and Uniqueness Theorem, continuity ensures that the function \(f(x, y)\) behaves nicely in a neighborhood around a given point.

For the given equation, \(f(x, y) = \frac{y+x}{y-x}\), continuity plays a crucial role. We found that this function isn't continuous along the line \(y = x\), where it becomes undefined. However, it remains continuous everywhere else in the \(xy\)-plane.

This means that we can only guarantee a unique solution for the differential equation in regions where \(y eq x\). By understanding where the function is continuous, we form a complete picture of where unique solutions to the differential equation can exist.