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(a) Use ten steps of Euler's method to approximate \(y-\) values for \(d y / d t=1 / t,\) starting at (1,0) and using \(\Delta t=0.1.\) (b) Using integration, solve the differential equation to find the exact value of \(y\) at the end of these ten steps. (c) Is your approximate value of \(y\) at the end of ten steps bigger or smaller than the exact value? Use a slope field to explain your answer.

Short Answer

Expert verified
The approximate value (0.74296) is greater than the exact value (0.69315). Euler's method overestimates in this scenario.

Step by step solution

01

Initializing Euler's Method

Euler's method approximates a solution for ordinary differential equations (ODEs) using an iterative approach. Start with the given point \((1,0)\) and define \( \Delta t = 0.1 \) as the step size. Our differential equation is \( \frac{dy}{dt} = \frac{1}{t} \).
02

First Iteration

Calculate the first approximation: \[ y_1 = y_0 + \Delta t \cdot \frac{1}{t_0} = 0 + 0.1 \cdot \frac{1}{1} = 0.1 \]. The new point is \((t_1, y_1) = (1.1, 0.1)\).
03

Second Iteration

Advance to the next point using Euler's formula:\[ y_2 = y_1 + \Delta t \cdot \frac{1}{t_1} = 0.1 + 0.1 \cdot \frac{1}{1.1} \approx 0.1909 \]. The new point is \((t_2, y_2) = (1.2, 0.1909)\).
04

Third Iteration

Continue with the iterative process:\[ y_3 = y_2 + \Delta t \cdot \frac{1}{t_2} = 0.1909 + 0.1 \cdot \frac{1}{1.2} \approx 0.27425 \]. The point becomes \((t_3, y_3) = (1.3, 0.27425)\).
05

Fourth Iteration

Compute the next step:\[ y_4 = y_3 + \Delta t \cdot \frac{1}{t_3} = 0.27425 + 0.1 \cdot \frac{1}{1.3} \approx 0.35182 \]. This gives the point \((t_4, y_4) = (1.4, 0.35182)\).
06

Fifth Iteration

Proceed to find:\[ y_5 = y_4 + \Delta t \cdot \frac{1}{t_4} = 0.35182 + 0.1 \cdot \frac{1}{1.4} \approx 0.42453 \]. We now have \((t_5, y_5) = (1.5, 0.42453)\).
07

Sixth Iteration

Calculate the next approximation:\[ y_6 = y_5 + \Delta t \cdot \frac{1}{t_5} = 0.42453 + 0.1 \cdot \frac{1}{1.5} \approx 0.49319 \]. The point is \((t_6, y_6) = (1.6, 0.49319)\).
08

Seventh Iteration

Continue calculating:\[ y_7 = y_6 + \Delta t \cdot \frac{1}{t_6} = 0.49319 + 0.1 \cdot \frac{1}{1.6} \approx 0.55869 \]. Now \((t_7, y_7) = (1.7, 0.55869)\).
09

Eighth Iteration

Compute the following step:\[ y_8 = y_7 + \Delta t \cdot \frac{1}{t_7} = 0.55869 + 0.1 \cdot \frac{1}{1.7} \approx 0.62194 \]. The point is \((t_8, y_8) = (1.8, 0.62194)\).
10

Ninth Iteration

Iterate once more:\[ y_9 = y_8 + \Delta t \cdot \frac{1}{t_8} = 0.62194 + 0.1 \cdot \frac{1}{1.8} \approx 0.68305 \]. This results in \((t_9, y_9) = (1.9, 0.68305)\).
11

Tenth Iteration

Finalize the approximation:\[ y_{10} = y_9 + \Delta t \cdot \frac{1}{t_9} = 0.68305 + 0.1 \cdot \frac{1}{1.9} \approx 0.74296 \]. The point is \((t_{10}, y_{10}) = (2.0, 0.74296)\).
12

Analytical Solution

Solve the differential equation analytically. Integrate both sides \( \int dy = \int \frac{1}{t} \, dt \), yielding \( y = \ln(t) + C \). With initial condition \((1,0)\), \ C = 0 \, hence \( y = \ln(t) \). At \( t = 2.0 \), \( y = \ln(2) \approx 0.69315 \).
13

Comparison and Conclusion

The approximate value after ten steps \(y_{10} \approx 0.74296 \) is greater than the exact value \( \ln(2) \approx 0.69315 \). Euler's method overestimates in this case, as the slope field shows decreasing slopes in the interval, causing the linear approximation to overshoot.

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Key Concepts

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

Ordinary Differential Equations
An Ordinary Differential Equation (ODE) is a relationship containing functions of only one independent variable and its derivatives. In mathematical terms, it is an equation with the form: \[ \frac{dy}{dx} = f(x, y) \] where \(f(x, y)\) is a known function. ODEs are commonly used to model and solve real-world problems where changes happen continuously. For example, they appear in physics to describe motion, in biology for population models, or in finance for growth rates.

The challenge with ODEs is that they often cannot be solved analytically, meaning we cannot always find a neat formula for the solution. Therefore, numerical methods like Euler's Method become important. Euler's Method allows us to find approximate solutions that give us insight into the behavior of the system described by the ODE.

Understanding how to approach ODEs involves recognizing the types and solutions available:
  • Separable Equations: Can be rearranged such that all terms with \(y\) are on one side and all terms with \(x\) on the other.
  • Linear Equations: Take on the form \(y' + P(x)y = Q(x)\).
  • Exact Equations: Satisfy a condition that allows for direct integration.
Euler’s Method specifically deals with cases where such analytical solutions aren't straightforward and provides step-by-step approximations instead.
Numerical Approximation
Numerical Approximation is a key technique used to find solutions to problems that cannot be solved exactly or analytically. In the context of ODEs, methods like Euler's Method provide an iterative way to approximate solutions.

Euler's Method works by taking a known solution point, using the derivative to estimate the next point, and repeating this process for a set number of steps. The method follows a simple formula: \[ y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n) \] where \(\Delta t\) is the step size and \(f(t_n, y_n)\) is the slope, calculated from the derivative at current point \((t_n, y_n)\).

The accuracy of the approximation depends on:
  • Step Size: Smaller step sizes generally yield more accurate results but require more computation.
  • Function Behavior: Sudden changes or non-linearity in the function can introduce larger errors.
  • Starting Conditions: Initial inaccuracies can propagate through subsequent calculations.
While approximating, one must be cautious of these factors to maintain an acceptable level of accuracy and reliability in the results.
Integration
Integration is the mathematical process of finding the integral of a function, often described as the inverse operation to differentiation. In solving differential equations analytically, integration helps us find a function \(y(x)\) whose derivative matches the function described by the ODE.

For instance, in solving the equation \(\frac{dy}{dt} = \frac{1}{t}\), the integration of both sides gives: \[ \int dy = \int \frac{1}{t} \, dt \] resulting in \[ y = \ln(t) + C \] where \(C\) is the constant of integration determined by initial conditions. In our specific problem, where \((t_0, y_0) = (1, 0)\), we find \(C = 0\), leading to \[ y = \ln(t) \].

The result of integration gives the exact solution to the differential equation over continuous intervals. Unlike numerical methods, which provide approximate solutions over discrete steps, integration provides a precise solution that is applicable for any value within the domain.
Slope Field Analysis
Slope Field Analysis provides a graphical method to visualize solutions of differential equations without solving them analytically. A slope field, or direction field, is created by drawing small lines—or strokes—that represent the slope \( \frac{dy}{dt} \) derived from the differential equation, at various points in the plane.

Each line in the slope field shows the direction in which a solution curve should move if it passed through that point. For example, given \( \frac{dy}{dt} = \frac{1}{t} \), lines would represent the reciprocal of \(t\) at each \((t, y)\) coordinate.

Slope fields help us understand
  • Solution Behavior: How solutions might trend over time.
  • Inflection Points: Where the curve changes direction.
  • Areas of Rapid Change: Regions where slope values are large.
In Euler’s Method context, observing a slope field can explain why an approximate solution might diverge from the exact one; particularly, overshooting occurs when slope field lines indicate where the true solution curves concave downward, leading the linear approximation of Euler’s Method to overshoot compared to the actual curve.

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Most popular questions from this chapter

A model for the population. \(P\), of carp in a landlocked lake at time \(t\) is given by the differential equation $$\frac{d P}{d t}=0.25 P(1-0.0004 P)$$ (a) What is the long-term equilibrium population of carp in the lake? (b) A census taken ten years ago found there were 1000 carp in the lake. Estimate the current population. (c) Under a plan to join the lake to a nearby river, the fish will be able to leave the lake. A net loss of \(10 \%\) of the carp each year is predicted, but the patterns of birth and death are not expected to change. Revise the differential equation to take this into account. Use the revised differential equation to predict the future development of the carp population.

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

Find the general solution to the given differential equation. $$y^{\prime \prime}+2 y^{\prime}+3 y=0$$

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(g(x)\) is periodic, then \(f(x)\) is also periodic.

Solve the boundary value problem. $$p^{\prime \prime}+2 p^{\prime}+2 p=0, \quad p(0)=0, \quad p(\pi / 2)=20$$

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