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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. \(y(0)=1 \quad y^{\prime}(t)=\sqrt{y} \quad y=(t / 2+1)^{2} \quad 0 \leq t \leq 1 \quad h=0.2\)

Short Answer

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

Step by step solution

01

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.
02

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. \( t=0.4, y_4 = 1.497 + 0.1 \times (1.497)^2 \approx 1.891 \)
03

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. \( y_4 = 1.537 + 0.05 (1.537^2 + (1.537 + 0.1 \times 1.537^2)^2) \approx 1.947 \)
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Continuous infusion of penicillin. Suppose a patient recovering from surgery is to be administered penicillin intravenously at a constant rate of 5 grams per hour. The patient's kidneys will remove penicillin at a rate proportional to the serum penicillin concentration. Let \(P(t)\) be the penicillin concentration \(t\) hours after infusion is begun. Then a simple model of penicillin pharmacokinetics is \(\begin{array}{l}\text { Net Rate of Increase } & \text { Clearance } & \text { Infusion }\end{array}\) $$P^{\prime}(t)=-K \times P(t)+5$$ $$\frac{\mathrm{gm}}{\mathrm{hr}} \quad \frac{1}{\mathrm{hr}} \times \operatorname{gm} \quad \frac{\mathrm{gm}}{\mathrm{hr}}$$ The proportionality constant, \(K,\) must have units \(\frac{1}{\mathrm{hr}}\) in order for the units on the equation to balance. We initially assume that \(K=2.5 \frac{1}{\mathrm{hr}}\) which is in the range of physiological reality. It is reasonable to assume that there was no penicillin in the patient at time \(t=0,\) so that \(P(0)=0\). a. Draw the phase plane for the differential equation $$P(0)=0 \quad P^{\prime}(t)=-2.5 P(t)+5$$ b. Find the equilibrium point of \(P^{\prime}=-2.5 P+5\). c. Is the equilibrium point stable? d. Show that the units of the equilibrium point are grams. e. Suppose the patient's kidneys are impaired and only operating at \(60 \%\) of normal. Then \(K=1.5\) instead of \(2.5 .\) What effect does this have on the equilibrium point.

Show that the variables are not separable in the equation $$\text { a. } \quad y^{\prime}(t)=\ln (t \times y) \quad \text { b. } \quad y^{\prime}(t)=\ln (t+y)$$

Some population scientists have argued that population density can get so low that reproduction will be less than natural attrition and the total population will be lost. Named the Allee effect. after W. C. Allee who wrote extensivelv about it \(^{6}\), this mav be a basis for arguing. for example that marine fishing of a certain species (Atlantic cod on Georges Bank, for example \({ }^{7}\) ee Paul Greenberg, "Four Fish, the Future of the Last Wild Food, The Penguin Press, 2010 . The model is a lot more complicated than we present here.) should be suspended, despite the presence of a small residual population. How should we modify the logistic differential equation, \(u^{\prime}=u(1-u),\) to incorporate such a threshold? Assume a fixed area and uniform density throughout the area and a threshold number, \(\epsilon .\) If the population number is less than \(\epsilon\) the population will decline; if the population number is more than \(\epsilon\) the population will increase. a. Modify the direction field for \(u^{\prime}=u(1-u)\) to account for the Allee effect. That is, a \(u, t\) plane, draw the line \(u=1\) and a threshold line \(u=\epsilon\) where \(\epsilon=0.1,\) say. Arrows below \(u=\epsilon\) should point downward; arrows between \(u=\epsilon\) and \(u=1\) should point upwards. Draw enough direction field arrows to indicate the paths of solutions for a threshold model. b. Draw the the logistic phase plane graph and the phase plane graphs for the following three candidates of a threshold logistic differential equation where \(\epsilon=0.1\). \(u^{\prime}=f(u)=u \times(1-u) \quad\) Logistic \(u^{\prime}=f_{1}(u)=u^{\frac{2}{3}} \times(u-\epsilon)^{\frac{1}{3}} \times(1-u) \quad\) Candidate 1 \(u^{\prime}=f_{2}(u)=u \times \frac{u-\epsilon}{u+\epsilon} \times(1-u)\) Candidate 2 \(u^{\prime}=f_{3}(u)=u \times(u-\epsilon) \times(1-u)\) Candidate 3

There is 'conventional wisdom' among SCUBA divers that if you are going to make a dive that involves two depths, 'do the deep part first'. This problem and the next explores rationale for that wisdom. To be concrete, assume that \(K=0.071 / \mathrm{min}\) which corresponds to approximately 10 minute half- life for the compartment \(((\ln 2) / 0.07=9.9 \mathrm{~min})\). a. Assume a diver (d1) descends immediately to 10 meters and stays there for 15 minutes, then descends to 30 meters and stays there for 10 minutes. Let $$d_{1}(t)=\left\\{\begin{array}{ll}10 & \text { for } 0 \leq t \leq 15 \\\30 & \text { for } 15

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