Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

A differential equation with initial condition and its analytic solution are shown. i. Show that the analytic solution satisfies the initial condition and the differential equation. ii. Use Euler's method and the trapezoid methods to approximate the solution to the differential equation on the interval shown and using the step size shown. iii. Plot the solution and the Euler's and trapezoid approximations on a single \(t, y\) plane. a. \(y(0)=1 \quad y^{\prime}(t)=y^{2} \quad y(t)=(1-t)^{-1} \quad 0 \leq t \leq 0.4 \quad h=0.1\) b. \(y(0)=2 \quad y^{\prime}(t)=-y^{2} \quad y(t)=(t+0.5)^{-1} \quad 0 \leq t \leq 0.4 \quad h=0.1\) c. \(y(0)=1 \quad y^{\prime}(t)=t \times y \quad y(t)=e^{t^{2} / 2} \quad 0 \leq t \leq 1 \quad h=0.2\) d. \(y(0)=1 \quad y^{\prime}(t)=\sqrt{y} \quad y=(t / 2+1)^{2} \quad 0 \leq t \leq 1 \quad h=0.2\)

Step-by-Step Solution

Answer

The analytic solution matches initial condition & diff. equation; Euler and Trapezoid give approximate values.

Step 1: Verify Initial Condition and Differential Equation for a

Given the differential equation is \( y'(t) = y^2 \) and the solution is \( y(t) = (1-t)^{-1} \). The initial condition is \( y(0)=1 \). Verify by plugging \( t=0 \) into the solution: \( y(0) = (1-0)^{-1} = 1 \). Next, check the differential equation: \( y(t) = (1-t)^{-1} \Rightarrow y'(t) = \frac{d}{dt}[(1-t)^{-1}] = (1-t)^{-2} \).Since \( y(t)^2 = [(1-t)^{-1}]^2 = (1-t)^{-2} \), both sides are equal, verifying the differential equation.

Step 2: Apply Euler's Method for Approximation a

Euler's method formula is: \( y_{n+1} = y_n + h \cdot f(t_n, y_n) \). Here, \( f(t,y) = y^2 \), \( h = 0.1 \), starting with \( y_0 = 1 \).1. \( t=0, y(0) = 1 \) 2. \( t=0.1, y_1 = 1 + 0.1 \times (1)^2 = 1.1 \)3. \( t=0.2, y_2 = 1.1 + 0.1 \times (1.1)^2 \approx 1.221 \)4. \( t=0.3, y_3 = 1.221 + 0.1 \times (1.221)^2 \approx 1.497 \)5. \( t=0.4, y_4 = 1.497 + 0.1 \times (1.497)^2 \approx 1.891 \)

Step 3: Apply Trapezoid Method for Approximation a

The Trapezoid method uses: \( y_{n+1} = y_n + \frac{h}{2} (f(t_n, y_n) + f(t_{n+1}, y^*)) \), where \( y^* = y_n + h \cdot f(t_n, y_n) \).For \( f(t,y) = y^2 \):1. \( y_1 = 1 + 0.05 (1^2 + (1+0.1\cdot 1)^2) = 1.11 \)2. \( y_2 = 1.11 + 0.05 (1.11^2 + (1.11 + 0.1 \cdot 1.11^2)^2) \approx 1.248 \)3. \( y_3 = 1.248 + 0.05 (1.248^2 + (1.248 + 0.1 \times 1.248^2)^2) \approx 1.537 \)4. \( y_4 = 1.537 + 0.05 (1.537^2 + (1.537 + 0.1 \times 1.537^2)^2) \approx 1.947 \)

Step 4: Plotting Solutions and Approximations a

Plot the following data points on a \(t, y\) plane:- Analytic Solution: \( y(t) = (1-t)^{-1} \) for \( t = 0, 0.1, 0.2, 0.3, 0.4 \). - Points: \( (0, 1), (0.1, 1.111), (0.2, 1.25), (0.3, 1.428), (0.4, 1.666) \).- Euler's Approximation: Points \( (0, 1), (0.1, 1.1), (0.2, 1.221), (0.3, 1.497), (0.4, 1.891) \).- Trapezoid Approximation: Points \( (0, 1), (0.1, 1.11), (0.2, 1.248), (0.3, 1.537), (0.4, 1.947) \).Compare the closeness of the approximations to the analytic solution curve.

Key Concepts

Euler's Method

Euler's Method is a simple yet powerful tool for solving differential equations when we don't have the luxury of an analytic solution. Essentially, it's a way to "step" through our equation, using a given initial condition. Here's how it works: given a slope or derivative, we calculate the next point based on our current point. This is defined by the formula: \[ y_{n+1} = y_n + h \cdot f(t_n, y_n) \]where:

• \( y_n \) is the current value,
• \( h \) is the step size,
• \( f(t_n, y_n) \) is the function describing the slope of the curve.

Euler's method works like drawing straight lines from point to point to create a picture of the curve. It's straightforward and gives a rough estimation of the curve, especially for smaller step sizes. However, because it's based on straight lines, it can miss the nuances of curves which is why the step size is crucial.

Trapezoid Method

The Trapezoid Method is another approach to approximate solutions for differential equations. It's a bit more sophisticated than Euler's method, as it takes into account the slope at both the current and next point to estimate the curve. This method is represented by:\[ y_{n+1} = y_n + \frac{h}{2} \cdot (f(t_n, y_n) + f(t_{n+1}, y^*)) \]where \( y^* = y_n + h \cdot f(t_n, y_n) \). The use of trapezoidal segments instead of straight lines makes it more accurate than Euler's method by averaging the slopes.

• It reduces error by considering the tangential direction at both endpoints of the interval.
• It's best used where higher precision is needed than what Euler's method can provide, but still maintains simplicity.

This approach helps in smoothing out the approximations and can produce a closer representation of the actual curve, particularly for nonlinear problems.

Differential Equations

Differential Equations are powerful tools used to model the way different variables change in relation to one another. They're the backbone of many mathematical models in physics, engineering, economics, and other sciences. At their core, they describe the rate of change of a quantity. For example, the heating of a pond by the sun can be modeled with such equations.The basic idea is that if you know how something changes, you can predict its future state. In our exercises, we tackle differential equations of the form \( y'(t) = y^2 \), where \( y^{\prime}(t) \) stands for the derivative of \( y \) with respect to \( t \). Such forms allow you to derive future values of \( y \) at different times \( t \) provided an initial condition is available.

• This requires solving either analytically by finding an exact solution or numerically via methods like Euler's or Trapezoid.

Analytic solutions give us a function that describes the system perfectly across all points, whereas numerical methods furnish approximate solutions.

Analytic Solution

Analytic Solutions to differential equations offer precise answers and are derived via algebraic manipulations. This means they describe the function perfectly and can be evaluated for any point. Deriving analytic solutions involves finding a function that satisfies the given differential equation across its entire domain.Take for instance the equation \( y'(t) = y^2 \). The analytic solution for this given the initial condition \( y(0) = 1 \) is \( y(t) = (1-t)^{-1} \). The task is to verify this by plugging the equation into the differential equation:

• The left-hand side is computed using derivatives: \( y'(t) = (1-t)^{-2} \).
• The right-hand side involves calculating \( y(t)^2 \), which must equal \( (1-t)^{-2} \).

Analytic solutions are not always obtainable, especially in complex systems. When it's possible, they provide a perfect model, enabling accurate predictions and insights across all potential inputs in their domain.