Question

Use a Taylor series to solve the equation $$ \frac{d y}{d x}+x y=0, \quad y(0)=1 $$ evaluating \(y(x)\) for \(x=0.0\) to \(0.5\) in steps of \(0.1\)

Short answer

\( y(x) = 1 - \frac{x^2}{2} \) and evaluated results: \( y(0.0) = 1 \), \( y(0.1) = 0.995 \), \( y(0.2) = 0.98 \), \( y(0.3) = 0.955 \), \( y(0.4) = 0.92 \), \( y(0.5) = 0.875 \).

Step-by-step solution

Understand the Differential Equation

The given differential equation is \ \( \frac{d y}{d x} + x y = 0 \) and the initial condition is \( y(0) = 1 \). The goal is to find an approximate solution \( y(x) \) using a Taylor series expansion around \( x = 0 \).

Write the Taylor Series Expansion

The Taylor series expansion of a function \( y(x) \) around \( x = 0 \) is given by \ \[ y(x) = y(0) + y'(0) x + \frac{y''(0)}{2!} x^2 + \frac{y'''(0)}{3!} x^3 + \ldots \]

Calculate the First Derivative \( y'(0) \)

From the differential equation \( \frac{d y}{d x} + x y = 0 \), substitute \( x = 0 \) and \( y(0) = 1 \). \ \( y'(0) + 0 \cdot 1 = 0 \implies y'(0) = 0 \)

Calculate the Second Derivative \( y''(0) \)

Differentiate both sides of the differential equation \( \frac{d y}{d x} + x y = 0 \) with respect to \( x \) to get: \ \( y'' + y + x y' = 0 \). Substitute \( x = 0 \), \( y(0) = 1 \), and \( y'(0) = 0 \): \ \( y''(0) + 1 + 0 = 0 \implies y''(0) = -1 \)

Calculate the Third Derivative \( y'''(0) \)

Differentiate again to get \( y''' + 2 y' + x y'' = 0 \). Substitute \( x = 0 \), \( y'(0) = 0 \), and \( y''(0) = -1 \): \ \( y'''(0) + 2 \cdot 0 + 0 \cdot (-1) = 0 \implies y'''(0) = 0 \)

Assemble the Taylor Series Expansion

Using the calculated derivatives, the Taylor series expansion of \( y(x) \) around \( x = 0 \) is: \ \[ y(x) = 1 + 0 \cdot x + \frac{-1}{2!} x^2 + 0 \cdot x^3 = 1 - \frac{x^2}{2} \]

Evaluate the Taylor Series at Specific Points

Evaluate \( y(x) \) at \( x = 0.0, 0.1, 0.2, 0.3, 0.4, 0.5 \) using \( y(x) = 1 - \frac{x^2}{2} \): \ \( y(0.0) = 1 \), \ \( y(0.1) = 1 - \frac{(0.1)^2}{2} = 0.995 \), \ \( y(0.2) = 1 - \frac{(0.2)^2}{2} = 0.98 \), \ \( y(0.3) = 1 - \frac{(0.3)^2}{2} = 0.955 \), \ \( y(0.4) = 1 - \frac{(0.4)^2}{2} = 0.92 \), \ \( y(0.5) = 1 - \frac{(0.5)^2}{2} = 0.875 \)

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These equations describe how a quantity changes over time or space. They are crucial in modeling various real-world phenomena such as motion, heat, and population growth.

For example, the equation \(\frac{d y}{d x} + x y = 0\) is a first-order linear differential equation. The goal is often to find the function y(x) that satisfies both the differential equation and initial conditions, like y(0) = 1 in this exercise.

Taylor Series

The Taylor series is a way to represent a function as an infinite sum of terms calculated from the function's derivatives at a specific point. Mathematically, for a function \(y(x)\), its Taylor series expansion around \(x = 0\) is given by:

\[ y(x) = y(0) + y'(0) x + \frac{y''(0)}{2!} x^2 + \frac{y'''(0)}{3!} x^3 + \ldots \]

In the exercise, the Taylor series is used to approximate the solution of the differential equation around \(x = 0\). Knowing the coefficients (derivatives at \(x = 0\)), we can build a series to estimate the function y(x).

Initial Value Problems

An initial value problem is a type of differential equation along with an initial condition that specifies the value of the solution at a particular point. Initial conditions help pinpoint the exact solution out of unlimited possibilities. For example, given the initial condition \( y(0) = 1 \), we can start finding specific values for y and its derivatives at x = 0.

This method prescribes how to project the solution over an interval. By solving the derivatives in sequence and applying the initial condition, we progressively build a Taylor series for the function.

Calculus

Calculus is the branch of mathematics focused on continuous change. It is essential for understanding the behavior of functions and solving differential equations. Two main tools in calculus are differentiation and integration.

Differentiation, which involves finding derivatives, helps determine how a function changes at any point. For instance, \( y'(0) \) is the slope of \(y(x)\) at \(x = 0\). Integration is the process of finding the area under a curve, essentially reversing differentiation.

In the given exercise, differentiation repeatedly finds higher-order derivatives \ (y'(0), y''(0), y'''(0), ...)\ ) necessary for building the Taylor series of the solution.