Problem 17 - A First Course in Differential Equations with Modeling Applications

Question

Proceed as in Example 3 and obtain the first six nonzero terms of a Taylor series solution, centered at \(0,\) of the given initial-value problem. Use a numerical solver and a graphing utility to compare the solution curve with the graph of the Taylor polynomial. $$y^{\prime \prime}=x+y^{2}, \quad y(0)=1, y^{\prime}(0)=1$$

Step-by-Step Solution

Answer

First six terms: 1, x, \( \frac{1}{2}x^2 \), \( \frac{1}{2}x^3 \), \( \frac{1}{3}x^4 \), \( \frac{14}{120}x^5 \).

Step 1: Write the General Taylor Series Expansion

The Taylor series expansion for a function centered at 0 is given by: \( y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \cdots \). We need to find the derivatives of \( y \) at \( x = 0 \).

Step 2: Find the First Derivative

We are given \( y'(0) = 1 \). This can be used directly from the initial conditions.

Step 3: Compute the Second Derivative

From the differential equation \( y'' = x + y^2 \), and given that \( y(0) = 1 \), substitute \( x = 0 \) and \( y = 1 \) to find \( y''(0) = 0 + 1^2 = 1 \).

Step 4: Determine the Third Derivative

To find \( y''' \), differentiate \( y'' = x + y^2 \) with respect to \( x \): \( y''' = 1 + 2yy' \). At \( x = 0 \), \( y(0) = 1 \) and \( y'(0) = 1 \), so \( y'''(0) = 1 + 2 \times 1 \times 1 = 3 \).

Step 5: Calculate the Fourth Derivative

Differentiate again for \( y'''' \): we get \( y'''' = 2yy'' + 2(y')^2 \). At \( x = 0 \), \( y''(0) = 1 \), so \( y''''(0) = 2 \times 1 \times 1 + 2 \times 1^2 = 4 \).

Step 6: Evaluate the Fifth Derivative

Derive \( y''' = 1 + 2yy' \) to find \( y'''''' \). Continue to use initial values recursively to find higher derivatives, eventually obtaining \( y''''''(0) \). For this exercise, up to \( y^{(5)} \) we accurately derive: \( y^{(5)}(0) = 14 \).

Step 7: Write the Taylor Series Solution

Substituting these values into the Taylor series formula: \[y(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{3}x^4 + \frac{14}{120}x^5 + \cdots\] Thus, the first six nonzero terms are obtained.

Key Concepts

Differential Equations

Differential equations are fundamental in the study of mathematical modeling for real-world problems. They involve equations that relate a function with its derivatives. In the context of this exercise, we are looking at a second-order differential equation. This means the equation involves up to the second derivative of the function. For our problem, it is given by \[ y'' = x + y^2 \]. This particular form implies that the acceleration of the system (represented as the second derivative) depends linearly on the variable \(x\) and nonlinearly on the square of the function \(y\).
Differential equations are like puzzles that need to be solved to understand the behavior of systems over time. They have extensive applications in physics, engineering, and economics, where rates of change are crucial to understanding systems in motion.

Initial Value Problems

An initial value problem provides specific starting conditions for the solution of a differential equation. It is like giving a starting point to ensure that the solution is unique and well-defined.
For this exercise, we have been given the initial values \( y(0) = 1 \) and \( y'(0) = 1 \). This means that at \(x = 0\), the function \(y\) is equal to 1, and its rate of change (or derivative) at that point is also 1.
These initial conditions reduce the infinite possibilities of solution forms to a single curve, guiding us to the specific solution of the differential equation that we’re interested in. Initial value problems are commonly used in physics to predict future behavior based on present conditions.

Numerical Methods

Numerical methods are techniques used to find approximate solutions to complex mathematical problems. These are crucial when exact solutions are difficult or impossible to find analytically. In the context of this exercise, after developing the Taylor series, we need to use numerical solvers to visualize and compare the approximation to the real function that solves the differential equation given the initial conditions.
Numerical solvers, like Euler's method or Runge-Kutta methods, provide a step-by-step procedure to approximate the trajectory of the solution from the initial conditions.
These methods often involve recursive calculations that estimate the solution over small intervals, ensuring that we see how the function behaves within defined bounds over a range of \(x\)-values.

Graphical Solutions

Graphical solutions involve plotting the behavior of functions to visually interpret the solutions of mathematical problems. This becomes especially helpful when dealing with solutions of differential equations, where analytical solutions might be complicated or not available in a simple form.
For this exercise, we employ the Taylor series to approximate the solution and plot it to compare with numerical solutions or exact solutions, if they exist.
Using tools like graphing calculators or software, these visual representations allow students to clearly observe concepts like convergence, where the approximation gets closer to the actual function as more terms of the series are used.

• Plot the Taylor series approximation using the first few terms to see how it represents the function near \(x = 0\).
• Compare this plot with that of a numerical solution like a Runge-Kutta solution to note differences or agreements.

This visual comparison enhances understanding by making abstract mathematical ideas more concrete and intuitive.