Problem 20 - Advanced Engineering Mathematics

Question

Suppose that \(y(x)\) denotes a solution of the initial-value problem \(y^{\prime}=x^{2}+y^{2}, y(1)=-1\) and that \(y(x)\) possesses at least a second derivative at \(x=1\). In some neighborhood of \(x=1\), use the DE to determine whether \(y(x)\) is increasing or decreasing, and whether the graph \(y(x)\) is concave up or concave down.

Step-by-Step Solution

Answer

At \( x=1 \), \( y(x) \) is increasing and concave down.

Step 1: Evaluate the derivative at the initial point

Substitute the given initial condition into the differential equation. Here, we have \( y' = x^2 + y^2 \). At \( x = 1 \) and \( y(1) = -1 \), so:\[ y'(1) = 1^2 + (-1)^2 = 1 + 1 = 2. \]Since \( y'(1) = 2 > 0 \), the function \( y(x) \) is increasing at \( x = 1 \).

Step 2: Find the second derivative of the DE

To examine concavity, we need the second derivative. Differentiate the DE: \( y' = x^2 + y^2 \), giving:\[ y'' = \frac{d}{dx}(x^2 + y^2) = 2x + 2yy'. \]

Step 3: Evaluate the second derivative at the initial point

Substitute \( x = 1 \), \( y = -1 \), and \( y' = 2 \) into the expression for \( y'' \):\[ y''(1) = 2 \times 1 + 2 \times (-1) \times 2 = 2 - 4 = -2. \]Since \( y''(1) = -2 < 0 \), \( y(x) \) is concave down at \( x = 1 \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They play a crucial role in modeling the behavior of dynamic systems. These equations can describe various real-world phenomena in fields such as physics, engineering, and biology. In the context of an initial value problem, a differential equation is used alongside an initial condition. This helps to determine a unique solution.
In our exercise, the differential equation is given by:

• \( y' = x^2 + y^2 \)

This equation signifies how the rate of change of \( y \), denoted as \( y' \), depends on both \( x \) and \( y \) themselves. Solving this type of equation usually involves integration or other analytical methods, with the aim of finding the function \( y(x) \).
Additionally, the initial condition \( y(1) = -1 \) provides a specific point on the solution curve, ensuring that our solution matches certain requirements exactly at that point. This is why the exercise specifies both the differential equation and the condition.

Concavity

Concavity describes the way a curve bends. It is linked with the second derivative of a function. Let's break this down a bit more.
For a function \( y(x) \):

• If the second derivative \( y''(x) > 0 \), the function is concave up at that point. Imagine a smiley face or the shape of a bowl—it curves upwards.
• If \( y''(x) < 0 \), the function is concave down at that point. This is similar to a sad face or an upside-down bowl—it curves downwards.

In our problem, at \( x = 1 \), we calculated the second derivative to be \( y''(1) = -2 \). Since this value is less than zero, it indicates that the function \( y(x) \) is concave down at this point. Understanding concavity helps in visualizing the behavior of curves and can be crucial for optimization and analysis in real-world applications.

Increasing and Decreasing Functions

Understanding when a function is increasing or decreasing is fundamental when analyzing graphs and behaviors of functions. This is closely related to the first derivative of a function.
If we consider a function \( y(x) \):

• It is increasing at point \( x \) if the derivative \( y'(x) > 0 \).
• It is decreasing at point \( x \) if \( y'(x) < 0 \).

In our initial value problem, we found that at the initial point \( x = 1 \), \( y'(1) = 2 \), which is greater than zero. Therefore, the function \( y(x) \) is increasing at this point. This implies that as \( x \) moves from this specified point, \( y(x) \) rises. Recognizing whether a function is increasing or decreasing can be essential for predicting trends and making calculations in practical scenarios.