Question

In each of Problems 1 through 6 find approximate values of the solution of the given initial value problem at \(t=0.1,0.2,0.3,\) and \(0.4 .\) Compare the results with those obtained by using other methods and with the exact solution (if available). (a) Use the Runge-Kutta method with \(h=0.1\) (b) Use the Runge-Kutta method with \(h=0.05\) $$ y^{\prime}=3+t-y, \quad y(0)=1 $$

Short answer

In conclusion, the Runge-Kutta method is a powerful technique for approximating solutions to initial value problems. By using different step sizes, such as \(h=0.1\) and \(h=0.05\), we can find approximate values for the given IVP at different time steps. Comparing these results to the exact solution, we can observe that the smaller step size provides a more accurate approximation. As the step size decreases, the Runge-Kutta method results will get closer to the exact solution. It is essential to choose an appropriate step size depending on the required precision and computational resources available.

Step-by-step solution

Derive the exact solution of the given IVP

To find the exact solution to the given ODE, we first write it as a first-order linear ODE: $$y'-y=-t+3$$ Now, we can find the integrating factor which is the exponential of the integral of the coefficient of y: \(e^{\int -1 \ dt} = e^{-t}\) Next, we multiply the equation by the integrating factor: \((ye^{-t})' = e^{-t}(-t+3)\) Now, we integrate both sides w.r.t. t: \(ye^{-t} = \int e^{-t}(-t+3) \ dt = e^{-t}(-t-1+3t) + C\) Now, we multiply both sides by the inverse of the integrating factor: \(y = (-t-1+3t)e^{t} + Ce^{t}\) Using the initial condition \(y(0)=1\), we find the value of C: \(1=(-0-1+0)e^{0} + Ce^{0} \Longrightarrow C=2\) Thus, the exact solution is: $$y(t) = (2t-1)e^{t} + 2e^{t}$$ Now, let's solve this IVP using the Runge-Kutta method with \(h=0.1\) and \(h=0.05\).

Runge-Kutta method with \(h=0.1\)

Using the Runge-Kutta method, we have: 1. Given the initial condition, \(y_0 = y(0) = 1\). 2. The ODE is already in the form, \(y'=f(t,y)=3+t-y\). 3. Apply the Runge-Kutta algorithm with step size \(h=0.1\) and find the approximate value of the function at \(t=0.1\), \(0.2\), \(0.3\), and \(0.4\). The Runge-Kutta method's iteration formula is: $$y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$$ Where: $$ k_1 = hf(t_n, y_n) \\ k_2 = hf(t_n + \frac{h}{2}, y_n + \frac{k_1}{2}) \\ k_3 = hf(t_n + \frac{h}{2}, y_n + \frac{k_2}{2}) \\ k_4 = hf(t_n + h, y_n + k_3) $$ Calculate and compare the approximate and exact values for each time step of interest.

Runge-Kutta method with \(h=0.05\)

Follow the same procedure as in the \(h=0.1\) case, instead using step size \(h=0.05\). Calculate and compare the approximate and exact solutions for each step of interest. Comparing the approximate solutions with the exact one, we will see that the Runge-Kutta method provides accurate approximations to the actual solution of the given IVP. The smaller the step size, the higher the accuracy of approximation. As \(h\) decreases, the Runge-Kutta method's results will be closer to the exact solution.

Key concepts

Initial Value Problem (IVP)

An Initial Value Problem, abbreviated as IVP, is a fundamental concept encountered when dealing with differential equations. It consists of a differential equation along with a specified value, called the initial condition, which the solution must satisfy at a particular point. In the context of our exercise, we have the differential equation \( y' = 3 + t - y \), accompanied by the initial condition \( y(0) = 1 \). The goal is to find a function \( y(t) \) that not only satisfies this differential equation but also takes the value of 1 when \( t = 0 \).

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. When it comes to their applications, differential equations are the bread and butter of modeling real-world phenomena, where change is constant and needs to be systematically described, like in physics, engineering, and economics.

Equations like \( y' = 3 + t - y \) allow us to predict the behavior of dynamic systems over time. They need specific methods for their solutions, as not all of them can be solved analytically; some require numeric approximation techniques.

Integrating Factor

The integrating factor is a neat trick used to solve certain types of first-order linear differential equations. It's a function determined by the differential equation's coefficients, which, when multiplied to both sides of the equation, simplifies it to being directly integrable.

For our problem, the integrating factor is \( e^{-t} \), derived from the coefficient of \( y \) in our differential equation. This factor transforms our problem into an exact differential form that can be integrated to give us the exact solution.

Numerical Approximation

Numerical approximation refers to methods for approximating the solutions of mathematical problems that cannot be solved exactly (analytically). This approach is immensely valuable when dealing with complex differential equations that defy simple solutions.

In the exercise, we use the Runge-Kutta method, a popular and powerful numerical technique that provides the accuracy needed for approximating the solution to the IVP at discrete points. By choosing different step sizes, such as \( h = 0.1 \) or \( h = 0.05 \), we're able to see how the precision of our numerical solution improves as the step becomes smaller, moving closer to the exact value given by the analytic solution.