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Chapter 27: Problem 15

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

Expert verified
The series is \(y(x) = 1 - \frac{x^2}{2}\). Evaluating at \(x \in \{0.0, 0.1, 0.2, 0.3, 0.4, 0.5\}\): \(y(0.0) = 1\), \(y(0.1) = 0.995\), \(y(0.2) = 0.98\), \(y(0.3) = 0.955\), \(y(0.4) = 0.92\), \(y(0.5) = 0.875\).

Step by step solution

01

- Taylor Series Expansion

Consider the Taylor series expansion for a function around a point, which is given by: \[ y(x) = y(0) + y'(0)x + \frac{y''(0)x^2}{2!} + \frac{y'''(0)x^3}{3!} + \dots \]For the given differential equation, we will compute the first few derivatives at \(x=0\) to construct the series up to the third-order term.
02

- Compute the initial condition

From the problem statement, we have \(y(0) = 1\). This gives us the value of the function at \(x=0\).
03

- Calculate the first derivative

The differential equation is \(\frac{dy}{dx} + xy = 0\). Solving for \(\frac{dy}{dx}\) gives us:\[ \frac{dy}{dx} = -xy \]At \(x=0\), we get \(\left.\frac{dy}{dx}\right|_{x=0} = -0 \times 1 = 0\). So, \(y'(0) = 0\).
04

- Calculate the second derivative

We need the second derivative \(\frac{d^2 y}{d x^2}\). Differentiating \(\frac{dy}{dx} = -xy\) with respect to \(x\) gives:\[ \frac{d^2 y}{d x^2} = -\left( y + x\frac{dy}{dx} \right) \]At \(x=0\), we get \(y(0) = 1\) and \(\left.\frac{dy}{dx}\right|_{x=0} = 0\), hence:\[ \left.\frac{d^2 y}{d x^2}\right|_{x=0} = -1 \]
05

- Calculate the third derivative

For the third derivative, we differentiate \( \frac{d^2 y}{d x^2} = -\left( y + x\frac{dy}{dx} \right) \).Using the product rule gives:\[ \frac{d^3 y}{d x^3} = -\left( \frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2 y}{d x^2} \right) \]At \(x=0\), with previous values substituted, we get:\[ \left.\frac{d^3 y}{d x^3}\right|_{x=0} = -\left(0 + 0 + 0 \times (-1) \right) = 0\]
06

- Construct the series expansion

Using the computed derivatives, the Taylor series up to the second order is:\[ y(x) \approx y(0) + y'(0)x + \frac{y''(0)x^2}{2!} + \frac{y'''(0)x^3}{3!} \]Substituting values, we get:\[ y(x) = 1 + 0 \cdot x - \frac{x^2}{2} + 0 \cdot \frac{x^3}{6} \]Thus, the series simplifies to:\[ y(x) = 1 - \frac{x^2}{2} \]
07

- Evaluate at given points

Evaluate the series at the given points:- For \(x = 0.0\): \(y(0.0) = 1 - \frac{(0.0)^2}{2} = 1\)- For \(x = 0.1\): \(y(0.1) = 1 - \frac{(0.1)^2}{2} = 1 - 0.005 = 0.995\)- For \(x = 0.2\): \(y(0.2) = 1 - \frac{(0.2)^2}{2} = 1 - 0.02 = 0.98\)- For \(x = 0.3\): \(y(0.3) = 1 - \frac{(0.3)^2}{2} = 1 - 0.045 = 0.955\)- For \(x = 0.4\): \(y(0.4) = 1 - \frac{(0.4)^2}{2} = 1 - 0.08 = 0.92\)- For \(x = 0.5\): \(y(0.5) = 1 - \frac{(0.5)^2}{2} = 1 - 0.125 = 0.875\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series Expansion
The Taylor series is a powerful mathematical tool for approximating functions. It expands a function into an infinite sum of terms calculated from the values of its derivatives at a single point. The general form of a Taylor series expansion of a function \( y(x) \) around a point \( x = a \) is:
\[ y(x) = y(a) + y'(a)(x - a) + \frac{y''(a)(x - a)^2}{2!} + \frac{y'''(a)(x - a)^3}{3!} + \dots \] Each term involves a higher-order derivative of the function and a power of \( x - a \). By using the Taylor series, we can approximate the function around the point \( a \), which in this exercise is particularly useful for solving differential equations.
Initial Value Problems
Initial value problems (IVPs) are a type of differential equation accompanied by a specific condition, known as the initial condition. These problems require finding a function that not only solves the differential equation but also satisfies the given initial condition.
In this exercise, the initial value problem is:
\[ \frac{d y}{d x} + x y = 0, \quad y(0) = 1 \] The initial condition \( y(0) = 1 \) specifies the value of the function at \( x = 0 \). This information is crucial for constructing the Taylor series, as it provides the necessary starting point or anchor for the solution.
Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. They describe how a function changes and are essential in various fields such as physics, engineering, and economics. The given differential equation in the exercise is:
\[ \frac{d y}{d x} + x y = 0 \] To solve it using the Taylor series, we need to compute the derivatives of the function at the initial point \( x = 0 \). These derivatives provide the coefficients for the Taylor series expansion, allowing us to approximate the solution near the initial point.
Higher-Order Derivatives
In solving differential equations using the Taylor series, higher-order derivatives play a crucial role. They determine the coefficients of the higher power terms in the Taylor series. In this exercise, we calculate up to the third-order derivative:
1. The first derivative:
\[ \frac{d y}{d x} = - x y \] Evaluating at \( x = 0 \) gives \( y'(0) = 0 \).

2. The second derivative:
\[ \frac{d^2 y}{d x^2} = - \left( y + x \frac{d y}{d x} \right) \] Evaluating at \( x = 0 \) gives \( y''(0) = -1 \).

3. The third derivative:
\[ \frac{d^3 y}{d x^3} = - \left( 2 \frac{d y}{d x} + x \frac{d^2 y}{d x^2} \right) \] Evaluating at \( x = 0 \) gives \( y'''(0) = 0 \).

Using these derivatives, we construct the Taylor series:
\[ y(x) \approx y(0) + y'(0)x + \frac{y''(0)x^2}{2!} + \frac{y'''(0)x^3}{3!} = 1 - \frac{x^2}{2} \] This series approximates the solution of the differential equation near \( x = 0 \).

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Most popular questions from this chapter

Write a computer program that would solve, for a range of values of \(\lambda\), the differential equation $$ \frac{d y}{d x}=\frac{1}{\sqrt{x^{2}+\lambda y^{2}}}, \quad y(0)=1 $$ using a third-order Runge-Kutta scheme. Consider the difficulties that might arise when \(\lambda<0\)

For some problems, numerical or algebraic experimentation may suggest the form of the complete solution. Consider the problem of numerically integrating the first-order wave equation $$ \frac{\partial u}{\partial t}+A \frac{\partial u}{\partial x}=0 $$ in which \(A\) is a positive constant. A finite difference scheme for this partial differential equation is $$ \frac{u(p, n+1)-u(p, n)}{\Delta t}+A \frac{u(p, n)-u(p-1, n)}{\Delta x}=0 $$ where \(x=p \Delta x\) and \(t=n \Delta t\), with \(p\) any integer and \(n\) a non- negative integer. The initial values are \(u(0,0)=1\) and \(u(p, 0)=0\) for \(p \neq 0\) (a) Carry the difference equation forward in time for two or three steps and attempt to identify the pattern of solution. Establish the criterion for the method to be numerically stable. (b) Suggest a general form for \(u(p, n)\), expressing it in generator function form, i.e. as ' \(u(p, n)\) is the coefficient of \(s^{p}\) in the expansion of \(G(n, s)\) '. (c) Using your form of solution (or that given in the answers!), obtain an explicit general expression for \(u(p, n)\) and verify it by direct substitution into the difference equation. (d) An analytic solution of the original PDE indicates that an initial disturbance propagates undistorted. Under what circumstances would the difference scheme reproduce that behaviour?

A more refined form of the Adams predictor-corrector method for solving the first-order differential equation $$ \frac{d y}{d x}=f(x, y) $$ is known as the Adams-Moulton-Bashforth scheme. At any stage (say the \(n\) th) in an \(N\) th-order scheme, the values of \(x\) and \(y\) at the previous \(N\) solution points are first used to predict the value of \(y_{n+1} .\) This approximate value of \(y\) at the next solution point, \(x_{n+1}\), denoted by \(\bar{y}_{n+1}\), is then used together with those at the previous \(N-1\) solution points to make a more refined (corrected) estimation of \(y\left(x_{n+1}\right)\). The calculational procedure for a third-order scheme is summarised by the two following two equations: \(\begin{array}{lrl}\bar{y}_{n+1}=y_{n}+h\left(a_{1} f_{n}+a_{2} f_{n-1}+a_{3} f_{n-2}\right) & \text { (predictor) } \\ y_{n+1} & =y_{n}+h\left(b_{1} f\left(x_{n+1}, \bar{y}_{n+1}\right)+b_{2} f_{n}+b_{3} f_{n-1}\right) & & \text { (corrector) }\end{array}\) (corrector). (a) Find Taylor series expansions for \(f_{n-1}\) and \(f_{n-2}\) in terms of the function \(f_{n}=f\left(x_{n}, y_{n}\right)\) and its derivatives at \(x_{n}\)(b) Substitute them into the predictor equation and, by making that expression for \(\bar{y}_{n+1}\) coincide with the true Taylor series for \(y_{n+1}\) up to order \(h^{3}\), establish simultaneous equations that determine the values of \(a_{1}, a_{2}\) and \(a_{3}\). (c) Find the Taylor series for \(f_{n+1}\) and substitute it and that for \(f_{n-1}\) into the corrector equation. Make the corrected prediction for \(y_{n+1}\) coincide with the true Taylor series by choosing the weights \(b_{1}, b_{2}\) and \(b_{3}\) appropriately. (d) The values of the numerical solution of the differential equation, $$ \frac{d y}{d x}=\frac{2(1+x) y+x^{3 / 2}}{2 x(1+x)} $$ at three values of \(x\) are given in the following table: \begin{tabular}{clll} \hline \hline\(x\) & \(0.1\) & \(0.2\) & \(0.3\) \\ \(y(x)\) & \(0.030628\) & \(0.084107\) & \(0.150328\) \\ \hline \hline \end{tabular} Use the above predictor-corrector scheme to find the value of \(y(0.4)\) and compare your answer with the accurate value, \(0.225577\).

Solve the following set of simultaneous equations using Gaussian elimination (including interchange where it is formally desirable): $$ \begin{aligned} x_{1}+3 x_{2}+4 x_{3}+2 x_{4} &=0 \\ 2 x_{1}+10 x_{2}-5 x_{3}+x_{4} &=6 \\ 4 x_{2}+3 x_{3}+3 x_{4} &=20 \\ -3 x_{1}+6 x_{2}+12 x_{3}-4 x_{4} &=16 \end{aligned} $$

Use the isocline approach to sketch the family of curves that satisfies the nonlinear first-order differential equation $$ \frac{d y}{d x}=\frac{a}{\sqrt{x^{2}+y^{2}}} $$
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