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Chapter 10: Problem 41

Find the value of \(X(1)\) for the initial-value problem $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x}{2 \sqrt{(} t+x)}, \quad x(0.5)=1 $$ using Euler's method with step size \(h=0.1\).

Short Answer

Expert verified
Using Euler's method, \( X(1) \approx 1.11503 \).

Step by step solution

01

Understand Euler's Method

Euler's method is a numerical technique to solve ordinary differential equations (ODEs) with a given initial value. It uses the formula: \[ x_{n+1} = x_n + h \cdot f(t_n, x_n) \]where \( h \) is the step size, \( f(t, x) \) is the derivative, \( x_n \) is the current value of \( x \), and \( x_{n+1} \) is the next value.
02

Identify Initial Values and Parameters

The initial condition is given: \( x(0.5) = 1 \). The step size \( h \) is 0.1. We need to find \( X(1) \) using Euler's method.
03

Calculate the First Iteration (From t=0.5 to t=0.6)

Using the given derivative, substitute \( t=0.5 \) and \( x=1 \) into:\[ \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{x}{2 \sqrt{t+x}} = \frac{1}{2 \sqrt{0.5+1}} = \frac{1}{2\sqrt{1.5}} \]Calculate the value:\[ \frac{1}{2\sqrt{1.5}} \approx 0.2887 \]Apply Euler's formula:\[ x_{0.6} = 1 + 0.1 \times 0.2887 = 1.02887 \]
04

Calculate the Second Iteration (From t=0.6 to t=0.7)

Now use \( t=0.6 \) and \( x=1.02887 \):\[ \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{1.02887}{2 \sqrt{0.6 + 1.02887}} = \frac{1.02887}{2 \sqrt{1.62887}} \]Calculate the value:\[ \frac{1.02887}{2\sqrt{1.62887}} \approx 0.2535 \]Apply Euler's formula:\[ x_{0.7} = 1.02887 + 0.1 \times 0.2535 = 1.05422 \]
05

Calculate the Third Iteration (From t=0.7 to t=0.8)

Use \( t=0.7 \) and \( x=1.05422 \):\[ \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{1.05422}{2 \sqrt{0.7 + 1.05422}} = \frac{1.05422}{2 \sqrt{1.75422}} \]Calculate the value:\[ \frac{1.05422}{2\sqrt{1.75422}} \approx 0.2266 \]Apply Euler's formula:\[ x_{0.8} = 1.05422 + 0.1 \times 0.2266 = 1.07688 \]
06

Calculate the Fourth Iteration (From t=0.8 to t=0.9)

Use \( t=0.8 \) and \( x=1.07688 \):\[ \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{1.07688}{2 \sqrt{0.8 + 1.07688}} = \frac{1.07688}{2 \sqrt{1.87688}} \]Calculate the value:\[ \frac{1.07688}{2\sqrt{1.87688}} \approx 0.2033 \]Apply Euler's formula:\[ x_{0.9} = 1.07688 + 0.1 \times 0.2033 = 1.09721 \]
07

Calculate the Fifth Iteration (From t=0.9 to t=1.0)

Finally, use \( t=0.9 \) and \( x=1.09721 \):\[ \frac{\mathrm{d} x}{\mathrm{~d} t} = \frac{1.09721}{2 \sqrt{0.9 + 1.09721}} = \frac{1.09721}{2 \sqrt{1.99721}} \]Calculate the value:\[ \frac{1.09721}{2\sqrt{1.99721}} \approx 0.1782 \]Apply Euler's formula:\[ x_{1.0} = 1.09721 + 0.1 \times 0.1782 = 1.11503 \]

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Key Concepts

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

Numerical Methods
Numerical methods are techniques used to approximate solutions for mathematical problems. These methods are especially useful for problems that can't be solved analytically, such as certain differential equations.
Euler's method is one such numerical method, used for solving ordinary differential equations (ODE) with initial values. It provides an approximate solution by taking small steps from a known starting point. Each step involves calculating the slope of the tangent to the curve at a particular point, then moving forward a small amount along that slope to find the next point.
In essence, numerical methods like Euler's offer a way to create a series of points that approximate the continuous solution to a problem. This helps bridge the gap between complex mathematical theories and practical problem-solving applications where exact solutions are difficult or impossible to find.
  • Useful in cases where analytical solutions are difficult
  • Applications include engineering, physics, and finance
  • Provides practical solutions using approximations
Ordinary Differential Equations
Ordinary differential equations, or ODEs, are equations involving derivatives of a function. These functions depend on one independent variable. In our exercise, we have an ODE pertaining to the derivative of a function with respect to time (denoted as \( \frac{\mathrm{d} x}{\mathrm{~d} t} \)).
The ODE describes how a quantity changes over time or across another variable. It focuses on the relationship between a function and its rates of change. Solving ODEs often requires understanding of initial conditions, as the same type of differential equation can have multiple solutions depending on these conditions.
  • Expresses the change of a quantity
  • Used in dynamics, biology, and economics
  • Solutions depend on initial conditions
Initial-Value Problem
An initial-value problem in mathematics refers to finding a function that satisfies a differential equation and meets a specified initial condition. It is called an initial-value problem because it starts with a specific value of the function at a particular point, which is used to find the rest of the function.
For example, if the value of a function at \( t = 0.5 \) is known to be 1, this becomes the starting point for applying Euler's method. From this value, Euler's method can be used iteratively to estimate values of the function at subsequent points.
Solving an initial-value problem is critical in many practical applications, where one needs to predict future behavior based on current conditions.
  • Provides a starting condition to solve ODEs
  • Essential for practical problem-solving in sciences
  • Links theory with real-world application scenarios

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Most popular questions from this chapter

A water treatment plant deals with a constant influx \(Q\) of polluted water with pollutant concentration \(s_{0}\). The treatment tank contains bacteria which consume the pollutant and protozoa which feed on the bacteria, thus keeping the bacteria from increasing too rapidly and overwhelming the system. If the concentration of the bacteria and the protozoa are denoted by \(b\) and \(p\) the system is governed by the differential equations $$ \begin{aligned} &\frac{\mathrm{d} s}{\mathrm{~d} t}=r\left(s_{0}-s\right)-\alpha m \frac{b s}{1+s} \\ &\frac{\mathrm{d} b}{\mathrm{~d} t}=-r b+m \frac{b s}{1+s}-\beta n \frac{b p}{1+p} \\ &\frac{\mathrm{d} p}{\mathrm{~d} t}=-r p+n \frac{b p}{1+p} \end{aligned} $$ Write a program to solve these equations numerically. Measurements have determined that the (biological) parameters \(\alpha, m, \beta\) and \(n\) have the values \(0.5,1.0,0.8\) and \(0.1\) respectively. The parameter \(r\) is a measure of the inflow rate of polluted water and \(s_{0}\) is the level of pollutant. Using the initial conditions \(s(0)=0, b(0)=0.2\) and \(p(0)=0.05\) determine the final steady level of pollutant if \(r=0.05\) and \(s_{0}=0.4\). What effect does doubling the inflow rate \((r)\) have?

The tread of a car tyre wears more rapidly as it becomes thinner. The tread- wear rate, measured in \(\mathrm{mm}\) per 10000 miles, may be modelled as $$ a+b(d-t)^{2} $$ where \(d\) is the initial tread depth, \(t\) is the current tread depth and \(a\) and \(b\) are constants. A tyre company takes measurements on a new design of tyre whose initial tread depth is \(8 \mathrm{~mm}\). When the tyre is new its wear rate is found to be \(1.03 \mathrm{~mm}\) per 10000 miles run and when the tread depth is reduced to \(4 \mathrm{~mm}\) the wear rate is \(3.43 \mathrm{~mm}\) per 10000 miles. Assuming that a tyre is discarded when the tread depth has been reduced to \(2 \mathrm{~mm}\) what is its estimated life?

A simple mass spring system, subject to light damping, is vibrating under the action of a periodic force \(F \cos p t .\) The equation of motion is $$ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+4 x=F \cos p t $$ where \(F\) and \(p\) are constants. Solve the differential equation for the displacement \(x(t) .\) Show that one part of the solution tends to zero as \(t \rightarrow \infty\) and show that the amplitude of the steady-state solution is $$ F\left[\left(4-p^{2}\right)^{2}+4 p^{2}\right]^{-1 / 2} $$ Hence show that resonance occurs when \(p=\sqrt{2}\).

Determine which members of the given sets are solutions of the following differential equations. Hence, in each case, write down the general solution of the differential equation. (a) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}=0 \quad\left\\{1, t, t^{2}, t^{3}, t^{4}, t^{5}, t^{6}\right\\}\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-p^{2} x=0 \quad\left\\{\mathrm{e}^{p t}, \mathrm{e}^{-p t}, \cos p t, \sin p t\right\\}\) (c) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-p^{4} x=0\) \(\left\\{\mathrm{e}^{p t}, \mathrm{e}^{-p t}, \cos p t, \sin p t, \cosh p t, \sinh p t\right\\}\) (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}=0\) \(\left\\{\cos 2 t, \sin 2 t, \mathrm{e}^{-2 t}, \mathrm{e}^{2 t}, t^{2}, t, 1\right\\}\) (e) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+4 \frac{\mathrm{d} x}{\mathrm{~d} t}=0\) \(\left\\{\cos 2 t, \sin 2 t, \mathrm{e}^{-2 t}, \mathrm{e}^{2 t}, t^{2}, t, 1\right\\}\) (f) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+x=0\) \(\left\\{\mathrm{e}^{t}, \mathrm{e}^{-t}, \mathrm{e}^{2 t}, \mathrm{e}^{-2 t}, t \mathrm{e}^{t}, t \mathrm{e}^{-t}, t \mathrm{e}^{2 t}, t \mathrm{e}^{-2 t}\right\\}\) (g) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-\frac{\mathrm{d} x}{\mathrm{~d} t}+x=0\) \(\left\\{\mathrm{e}^{t}, \mathrm{e}^{-t}, \mathrm{e}^{2 t}, \mathrm{e}^{-2 t}, t \mathrm{e}^{t}, t \mathrm{e}^{-t}, t \mathrm{e}^{2 t}, t \mathrm{e}^{-2 t}\right\\}\)

Show that the characteristic equation of the differential equation $$ \frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-9 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-11 \frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=0 $$ is $$ \left(m^{3}+3 m^{2}+3 m+1\right)(m-4)=0 $$ and hence find the general solution of the equation.
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