Question

In each exercise, (a) Solve the initial value problem analytically, using an appropriate solution technique. (b) For the given initial value problem, write the Heun's method algorithm, $$ y_{n+1}=y_{n}+\frac{h}{2}\left[f\left(t_{n}, y_{n}\right)+f\left(t_{n+1}, y_{n}+h f\left(t_{n}, y_{n}\right)\right)\right] . $$ (c) For the given initial value problem, write the modified Euler's method algorithm, $$ y_{n+1}=y_{n}+h f\left(t_{n}+\frac{h}{2}, y_{n}+\frac{h}{2} f\left(t_{n}, y_{n}\right)\right) . $$ (d) Use a step size \(h=0.1\). Compute the first three approximations, \(y_{1}, y_{2}, y_{3}\), using the method in part (b). (e) Use a step size \(h=0.1\). Compute the first three approximations, \(y_{1}, y_{2}, y_{3}\), using the method in part (c). (f) For comparison, calculate and list the exact solution values, \(y\left(t_{1}\right), y\left(t_{2}\right), y\left(t_{3}\right)\). \(y^{\prime}=-y+t, \quad y(0)=0\)

Short answer

In summary, we have the following values for the two numerical methods and the exact solution: 1. Heun's method: \(y_1 = 0.005\), \(y_2 = 0.0195\), and \(y_3 = 0.042075\). 2. Modified Euler's method: \(y_1 = 0.005\), \(y_2 = 0.019025\), and \(y_3 = 0.041173875\). 3. Exact solution: \(y(0.1) = 0.1\), \(y(0.2) = 0.2\), and \(y(0.3) =0.3\). Both Heun's method and modified Euler's method provide approximate values for the given IVP. Although these values are not exact, they can still be helpful in understanding the overall behavior of the solution. In this case, it appears that Heun's method might produce slightly more accurate solutions than the modified Euler's method, but it should be noted that both methods exhibit some deviation from the exact solution values.

Step-by-step solution

Separation of variables method

Start by separating the variables and integrating both sides:$$\frac{dy}{dt}=-y+t.$$Divide both sides by \((-y + t)\) and integrate:$$\int \frac{dy}{-y + t} = \int dt.$$Now, let \(u = -y + t\) which means \(du = -dy\), thus$$\int \frac{-du}{u} = \int dt.\\\ln(|u|) = -t + C \\\ln(|-y+t|)=-t+C.$$To simplify further, we have$$-y+t=Ce^{-t}.$$We are given that when \(t=0\), \(y=0\). Using this information, we can solve for \(C\):$$0=-0+Ce^{-0}\Rightarrow C=0.$$So the exact solution to the IVP is$$y(t)=-y+t=0.$$So \(y(t)=t\). #Step 2: Write the Heun's method algorithm# For the given IVP, we can write the Heun's method algorithm by plugging our function into Heun's method formula. We have \(f(t,y)=-y+t\), then$$y_{n+1}=y_n+\frac{h}{2}\left[f\left(t_n, y_n\right)+f\left(t_{n+1}, y_n+ h f\left(t_n, y_n\right)\right)\right].$$ #Step 3: Write the modified Euler's method algorithm# For the given IVP, we can write the modified Euler's method algorithm by plugging our function into the modified Euler's method formula. We have \(f(t,y)=-y+t\), then$$y_{n+1}=y_n+ h f\left(t_n+\frac{h}{2}, y_n+\frac{h}{2} f\left(t_n, y_n\right)\right).$$ #Step 4: Compute the first three approximations using Heun's method# Let's use a step size \(h=0.1\), and compute the first three approximations \(y_1, y_2, y_3\) using Heun's method algorithm.

Calculate \(y_1, y_2, y_3\) using Heun's method

We are given the initial values \(y_0 = 0\) and \(t_0 = 0\). Let's calculate the following approximations using the Heun's method algorithm: 1. \(t_1 = 0.1\), \(y_1 = y_0 + \frac{0.1}{2}[f(0, 0) + f(0.1, 0 + 0.1*f(0, 0))] = 0 + \frac{0.1}{2}[0 + 0.1] = 0.005\). 2. \(t_2 = 0.2\), \(y_2 = y_1 + \frac{0.1}{2}[f(0.1, 0.005) + f(0.2, 0.005 + 0.1*f(0.1, 0.005))] = 0.005 + \frac{0.1}{2}[0.095 + 0.185] = 0.0195\). 3. \(t_3 = 0.3\), \(y_3 = y_2 + \frac{0.1}{2}[f(0.2, 0.0195) + f(0.3, 0.0195 + 0.1*f(0.2, 0.0195))] = 0.0195 + \frac{0.1}{2}[0.1805 + 0.262] = 0.042075\). So, using Heun's method, we have \(y_1 = 0.005\), \(y_2 = 0.0195\), and \(y_3 = 0.042075\). #Step 5: Compute the first three approximations using modified Euler's method# Now let's use the same step size \(h=0.1\) and compute the first three approximations \(y_1, y_2, y_3\) using the modified Euler's method algorithm.

Calculate \(y_1, y_2, y_3\) using modified Euler's method

We will use the initial values \(y_0 = 0\) and \(t_0 = 0\). Let's calculate the following approximations using the modified Euler's method algorithm: 1. \(t_1 = 0.1\), \(y_1 = y_0 + 0.1*f(0+\frac{0.1}{2}, 0+\frac{0.1}{2}*f(0, 0)) = 0 + 0.1*f(0.05, 0.05) = 0.1 * 0.05 = 0.005\). 2. \(t_2 = 0.2\), \(y_2 = y_1 + 0.1*f(0.1+\frac{0.1}{2}, 0.005+\frac{0.1}{2}*f(0.1, 0.005)) = 0.005 + 0.1*f(0.15, 0.00975) = 0.005 + 0.1 * 0.14025 = 0.019025\). 3. \(t_3 = 0.3\), \(y_3 = y_2 + 0.1*f(0.2+\frac{0.1}{2}, 0.019025+\frac{0.1}{2}*f(0.2, 0.019025)) = 0.019025 + 0.1*f(0.25, 0.02851375) = 0.019025 + 0.1 * 0.22148625 = 0.041173875\). So, using modified Euler's method, we have \(y_1 = 0.005\), \(y_2 = 0.019025\), and \(y_3 = 0.041173875\). #Step 6: Calculate and list the exact solution values# For comparison, let's calculate and list the exact solution values at \(t_1, t_2, t_3\).

Calculate exact solution values

Since the exact solution is \(y(t) = t\), we can write: 1. \(y(t_1) = y(0.1) = 0.1\). 2. \(y(t_2) = y(0.2) = 0.2\). 3. \(y(t_3) = y(0.3) = 0.3\). So the exact solution values are \(y(0.1) = 0.1\), \(y(0.2) = 0.2\), and \(y(0.3) =0.3\).

Key concepts

Understanding Initial Value Problems

An initial value problem (IVP) in the context of differential equations is a problem where you are given a function that models a rate of change and an initial condition. This function usually takes the form of a derivative, such as \( y'(t)\), which embodies how a certain quantity, \( y\), changes over the variable \( t\). The starting point, known as the initial condition, is a specific value at a specific time, such as \( y(0) = 0 \).

The goal with an IVP is to find the function \( y(t) \) that satisfies both the differential equation and the initial condition. Solving an IVP analytically can involve techniques such as separation of variables, which leverages the relationship between the differential equation and its integral to discover the actual function that aligns with the given conditions.

Exploring Heun's Method

Heun's method, also referred to as the improved or modified Euler's method, is a technique for numerically approximating the solutions to differential equations. It is a predictor-corrector method that starts with an initial estimate of the solution, commonly derived from the Euler's method, and then refines this estimate to enhance accuracy.

The process entails an initial 'predict' step, which predicts the value of \( y \) at the next point, and a 'correct' step that adjusts this prediction by averaging the slope at the current point and the slope at the predicted next point. Here's a concise formula representing Heun's method:
\[ y_{n+1} = y_{n} + \frac{h}{2}\left[f\left(t_{n}, y_{n}\right) + f\left(t_{n+1}, y_{n} + h f\left(t_{n}, y_{n}\right)\right)\right] \]This approach helps to achieve a better approximation as compared to using Euler's method alone, as it accounts for changes in the slope that Euler's method might miss.

Numerical Approximation Techniques

Numerical approximation encompasses a variety of algorithms and techniques used to find approximate solutions to problems that may not be solvable analytically. When dealing with differential equations, these methods are vital as they provide a way to estimate the values of the unknown function at discrete points.

Methods such as Heun's and Euler's take a step-by-step approach, where the equation is solved iteratively over small increments, known as 'steps', beginning from the initial condition. By specifying the size of these steps with a parameter \(h\), one can control the precision of the approximation; a smaller \(h\) typically results in a more accurate result, but requires more computational effort. Through these approaches, computational tools are able to model complex systems and scenarios in fields such as physics, engineering, and economics.

The Quest for the Exact Solution

While numerical methods are invaluable for approximating solutions, finding the exact solution to an IVP or any differential equation reveals the precise behavior of the system under consideration. An exact solution is a closed-form expression, such as \( y(t) = t\) for our example, that satisfies the differential equation for all values within a certain domain and fulfills the initial condition.

Exact solutions can often be obtained for simpler, linear differential equations or those that are separable. However, many real-world problems lead to complex differential equations that resist such analytical solutions, hence, the reliance on numerical approximations. Nonetheless, when an exact solution is obtainable and known, it serves as a benchmark to assess the accuracy of the numerical methods, providing insights into the reliability and potential improvements for these approximative algorithms.