Question

In each exercise, (a) Write the Euler's method iteration \(y_{k+1}=y_{k}+h f\left(t_{k}, y_{k}\right)\) for the given problem. Also, identify the values \(t_{0}\) and \(y_{0}\). (b) Using step size \(h=0.1\), compute the approximations \(y_{1}, y_{2}\), and \(y_{3}\). (c) Solve the given problem analytically. (d) Using the results from (b) and (c), tabulate the errors \(e_{k}=y\left(t_{k}\right)-y_{k}\) for \(k=1,2,3\). \(y^{\prime}=-t y, \quad y(0)=1\)

Short answer

Answer: The error for k = 2 is approximately 0.01960.

Step-by-step solution

Part (a) - Euler's method formula, t0, and y0

Given the initial value problem: \(y'(t) = -ty(t)\) with the initial condition \(y(0)=1\). In the Euler's method iteration form, we have the following equation: \(y_{k+1} = y_{k}+h\cdot f(t_{k}, y_{k})\). For this problem, \(f(t_{k}, y_{k}) = -t_{k}y_{k}\). Replacing this into the iteration formula, we get: \(y_{k+1} = y_{k} - h\cdot t_{k}y_{k}\). For this problem, \(t_0 = 0\) and \(y_0 = 1\) are the initial values.

Part (b) - Computing y1, y2, and y3

Using the step size \(h=0.1\), we will compute \(y_{1}, y_{2}\), and \(y_{3}\) using the Euler's method formula: For \(k = 0\), we have: \(y_1 = y_0 - h\cdot t_0y_0 = 1 - 0.1\cdot 0\cdot 1 = 1\) For \(k = 1\), we have: \(y_2 = y_1 - h\cdot t_1y_1 = 1 - 0.1\cdot 0.1\cdot 1 = 0.99\) For \(k = 2\), we have: \(y_3 = y_2 - h\cdot t_2y_2 = 0.99 - 0.1\cdot 0.2\cdot 0.99 = 0.9798\) Thus, the approximations are: \(y_1 = 1\), \(y_2 = 0.99\), and \(y_3 = 0.9798\).

Part (c) - Analytical solution

To solve the given problem analytically, we consider the first-order linear differential equation \(y'(t) = -ty(t)\). This is a first-order homogeneous equation, and we can use the integrating factor method to solve it. The integrating factor is given by \(I(t) = e^{\int -t \; dt} = e^{-\frac{1}{2}t^2}\). Now, we multiply both sides of the ODE by the integrating factor: \(e^{-\frac{1}{2}t^2}y'(t) + te^{-\frac{1}{2}t^2}y(t) = 0\) The left-hand side is the derivative of the product \(y(t)e^{-\frac{1}{2}t^2}\), so we have: \((y(t)e^{-\frac{1}{2}t^2})' = 0\) Integrating both sides, we get: \(y(t)e^{-\frac{1}{2}t^2} = C\) To find the constant C, we use the initial condition \(y(0) = 1\): \(1 = Ce^{0}\), which implies \(C = 1\) So, the analytical solution is: \(y(t) = e^{\frac{1}{2}t^2}\)

Part (d) - Tabulation of errors

Now, we will calculate the errors \(e_k = y(t_k) - y_k\) for \(k=1,2,3\). For \(k = 1\) \((t_1 = 0.1)\): \(e_1 = e^{\frac{1}{2}(0.1)^2} - 1 \approx 0.00503\) For \(k = 2\) \((t_2 = 0.2)\): \(e_2 = e^{\frac{1}{2}(0.2)^2} - 0.99 \approx 0.019602\) For \(k = 3\) \((t_3 = 0.3)\): \(e_3 = e^{\frac{1}{2}(0.3)^2} - 0.9798 \approx 0.042974\) The errors are tabulated below: | k | \(t_k\) | \(y_k\) | \(y(t_k)\) | \(e_k\) | |---|-------|--------|-------------------|--------------| | 1 | 0.1 | 1 | \(\approx 1.00503\) | \(\approx 0.00503\) | | 2 | 0.2 | 0.99 | \(\approx 1.00960\) | \(\approx 0.01960\) | | 3 | 0.3 | 0.9798 | \(\approx 1.02277\) | \(\approx 0.04297\) |

Key concepts

Differential Equations

Differential equations are mathematical equations that describe how certain quantities change over time or space. They form the foundation of modeling any system where there is a continuous change, such as physics, engineering, economics, and biology. A first-order differential equation involves the first derivative of a variable — in our exercise, the equation is ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ( ... ) which relates the rate of change of the function y(t) with respect to the independent variable t. Solutions to these equations can predict the behavior of dynamic systems, and understanding this allows us to forecast future events in natural phenomena and create engineering systems that behave predictably. In our exercise, we look at how Euler's method helps solve such equations when an analytical solution might be complex or unknown.

Numerical Analysis

Numerical analysis is the study of algorithms that use numerical approximation for the problems of mathematical analysis. It's used by engineers and scientists to get approximate solutions to complex mathematical problems. Euler's method, a fundamental aspect of numerical analysis, is a simple yet powerful technique that allows for the approximate solution of ordinary differential equations (ODEs). It is particularly useful when an analytical solution is either unavailable or difficult to obtain. In the exercise at hand, Euler's method takes a known initial value and uses the ODE to find successive values, hence constructing a piecewise linear approximation to the solution curve, step by step.

Initial Value Problem

An initial value problem in the context of differential equations is where we are given a function that describes the rate of change of some quantity, and we're to find the actual quantity at a subsequent time given its initial value. These problems are indispensable for making predictions based on a model. Our exercise illustrates this perfectly: we're given the differential equation ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ( ... ) and are to find the solution curve starting at the point where ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ( ... ) t=0 and y(0)=1. The goal is to predict y(t) at other points down the line.

Analytical Solution

An analytical solution to a differential equation is an exact expression that describes the behavior of the system and is obtained by methods of calculus. These solutions give us a formula that can predict an infinite number of points along a function. For the equation in our exercise, the analytical solution is found using an integrating factor. While Euler's method gives us an approximation, this solution gives us the precise values of y(t). Recognizing the difference between such exact and approximate solutions is crucial in understanding the limitations and benefits of computational techniques like Euler's method.