Question

Approximate the solution to the given differential equation with a degree 4 Maclaurin polynomial. $$y^{\prime}=y, \quad y(0)=1$$

Short answer

The polynomial approximation is \( y(x) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} \).

Step-by-step solution

Write the Differential Equation and Initial Condition

The given differential equation is \( y' = y \) with the initial condition \( y(0) = 1 \). We're tasked to approximate its solution using a degree 4 Maclaurin polynomial.

Write the General Form of the Maclaurin Series

A Maclaurin series for a function \( y(x) \) expanded around \( x=0 \) is given by \( y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \frac{y^{(4)}(0)}{4!}x^4 + \ldots \), where each derivative is evaluated at \( x = 0 \).

Find the First Derivative at \( x=0 \)

From the equation \( y' = y \), the first derivative at \( x=0 \) is \( y'(0) = y(0) = 1 \).

Find Higher Order Derivatives at \( x=0 \)

For each derivative \( y^{(n)} = y \), thus \( y'' = y', \; y''' = y'', \; y^{(4)} = y''' \). Since \( y' = y \) for all orders, we apply this recursively: \( y'' = 1 \), \( y''' = 1 \), and \( y^{(4)} = 1 \).

Substitute Into Maclaurin Series

Substitute \( y(0) = 1 \), \( y'(0) = 1 \), \( y''(0) = 1 \), \( y'''(0) = 1 \), and \( y^{(4)}(0) = 1 \) into the series: \[ y(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} \] which simplifies to \[ y(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} \].

Write the Final Approximation

The degree 4 Maclaurin polynomial approximation for the solution of the differential equation is \[ y(x) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} \].

Key concepts

Differential Equations

Differential equations are a fundamental concept in mathematics and are essential for modeling many real-world phenomena. They relate a function with its derivatives, describing how a certain quantity changes over time or space.
In our problem, the differential equation is given as \( y' = y \), representing an exponential growth process where the rate of change of \( y \) is proportional to the value of \( y \) itself. This type of first-order differential equation is common in problems involving growth, decay, and resource dynamics.
The goal here is to approximate the solution to this differential equation. Initially, the problem provides an initial condition \( y(0) = 1 \), which means that at time \( t = 0 \), the value of \( y \) is 1. This critical piece of information enables us to determine the specific solution to the equation that fits the given scenario.

Polynomial Approximation

Polynomial approximation is a technique used in mathematics to approximate more complex functions using simpler polynomial expressions. Using polynomial approximation can make calculations easier by providing a simple formula to estimate the value of a function at any given point.
In this exercise, we employed a Maclaurin series, which is a specific type of polynomial approximation around \( x=0 \). By expanding the function using derivatives at \( x=0 \), we construct a polynomial that represents the function up to a certain degree, called the degree of the polynomial.
The Maclaurin series for our function \( y(x) \) is constructed using the formula:

• \( y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \ldots + \frac{y^{(n)}(0)}{n!}x^n \)

For a degree 4 approximation, we consider terms up to \( x^4 \). By substituting the evaluated derivatives at \( x=0 \), we derive the polynomial \( y(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} \). This polynomial provides an excellent approximation of our original function within the range close to \( x=0 \).

Initial Value Problems

An initial value problem involves a differential equation along with an initial condition. This condition provides the starting point of the function, allowing us to find a particular solution to the differential equation.
In this exercise, the initial condition given is \( y(0) = 1 \). This defines the value of the solution at the point \( x = 0 \). Based on this starting value, we can uniquely determine the corresponding Maclaurin series expansion for the solution of the differential equation.
Solving an initial value problem typically involves:

• Identifying the differential equation.
• Using the initial condition to find the constant of integration, if applicable.
• Applying techniques like series expansion to approximate the solution where necessary.

The process of using an initial value, in this case, ensures that the series developed gives an accurate depiction of the function’s behavior around \( x=0 \). This kind of problem-solving is vital in fields like physics and engineering, where specific initial conditions often dictate outcomes.