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For the following initial value problems, compute the first two approximations \(u_{1}\) and \(u_{2}\) given by Euler's method using the given time step. $$y^{\prime}(t)=2 y, y(0)=2 ; \Delta t=0.5$$

Short Answer

Expert verified
Answer: The first two approximations are \(u_1 = 4\) and \(u_2 = 8\).

Step by step solution

01

Identify the given problem and initial condition

We have the initial value problem \(y^{\prime}(t) = 2y\) with the initial condition \(y(0) = 2\). The time step given is \(\Delta t = 0.5\).
02

Identify the function f(t, y)

The function to be used in Euler's method formula is found on the right side of the differential equation. In this case, it's \(f(t, y) = 2y\).
03

Calculate the first approximation \(u_1\)

Using Euler's method formula, let's calculate the first approximation \(u_1\). Here, \(t_0 = 0\), \(u_0 = y(0) = 2\), \(\Delta t = 0.5\). The formula is as follows: $$u_{n+1} = u_n + \Delta t \cdot f(t_n, u_n)$$ Replace the variables for the first approximation: $$u_1 = u_0 + \Delta t \cdot f(t_0, u_0) = 2 + 0.5 \cdot 2 \cdot 2$$ Compute the result: $$u_1 = 2 + 2 = 4$$
04

Calculate the second approximation \(u_2\)

Now, let's calculate the second approximation \(u_2\). Here, \(t_1 = 0.5, u_1 = 4, \Delta t= 0.5\). Use the same formula from the previous step: $$u_{n+1} = u_n + \Delta t \cdot f(t_n, u_n)$$ Replace the variables for the second approximation: $$u_2 = u_1 + \Delta t \cdot f(t_1, u_1) = 4 + 0.5 \cdot 2 \cdot 4$$ Compute the result: $$u_2 = 4 + 4 = 8$$
05

Conclusion

The first two approximations using Euler's method for the initial value problem \(y^{\prime}(t)=2y, y(0)=2 ; \Delta t=0.5\) are: $$u_1 = 4$$ $$u_2 = 8$$

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Key Concepts

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

Initial Value Problem
In mathematics, an initial value problem lays the foundation for needing solutions to differential equations. Essentially, it involves determining a function from its derivative (or rate of change) and an initial condition.
This initial condition typically specifies the value of the function at a given point, often called the starting point. For example, in this exercise, the initial condition is given as \( y(0) = 2 \).
This means that at time \( t = 0 \), the value of the function \( y \) is 2. Solving an initial value problem requires not just finding a family of solutions to the differential equation, but also picking out the one solution that fits the given initial condition.
  • Initial Value: Specifies where the function starts, such as \( y(0) = 2 \).
  • Objective: Find the specific function solution that satisfies both the differential equation and initial condition.
Differential Equations
Differential equations play a crucial role in modeling situations where quantities change over time. They involve equations that relate a function to its derivatives, expressing how the function changes. In this particular problem, the differential equation is \( y^{\prime}(t) = 2y \).
This equation suggests a relationship between the rate of change of \( y \), denoted as \( y^{\prime}(t) \), and the current value of \( y \) itself. Such equations are common in real-world scenarios, ranging from population growth to the decay of radioactive substances.
  • Significance: Illustrates how a function evolves over time.
  • Solution: Finding a function or set of functions that satisfy the equation throughout its domain.
The challenge with differentials is that while some have analytic solutions, others require numerical techniques like Euler's method.
Numerical Approximation
Numerical approximation, like Euler's Method, is vital when solving differential equations that don't have straightforward solutions. Euler’s Method allows us to calculate function values at discrete steps by finding slopes at points and moving a small step (\( \Delta t \)) along the slope.
In this example, we used a step size of \( \Delta t = 0.5 \) to approximate values \( u_1 \) and \( u_2 \). The method doesn't provide perfect solutions but offers a close enough approximation to understand how the function behaves over an interval.
  • Step Size \( \Delta t \): Determines precision and is the interval between successive points.
  • Process: Estimate the next point using current estimate and slope.
Euler's method is straightforward but might require smaller step sizes for better accuracy, yet larger sizes are computationally cheaper.

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Most popular questions from this chapter

Solve the initial value problem $$M^{\prime}(t)=-r M \ln \left(\frac{M}{K}\right), \quad M(0)=M_{0}$$ with arbitrary positive values of \(r, K,\) and \(M_{0}\)

Use a calculator or computer program to carry out the following steps. a. Approximate the value of \(y(T)\) using Euler's method with the given time step on the interval \([0, T]\). b. Using the exact solution (also given), find the error in the approximation to \(y(T)\) (only at the right endpoint of the time interval). c. Repeating parts (a) and (b) using half the time step used in those calculations, again find an approximation to \(y(T)\). d. Compare the errors in the approximations to \(y(T)\). $$\begin{array}{l}y^{\prime}(t)=t-y, y(0)=4 ; \Delta t=0.2, T=4; \\\y(t)=5 e^{-t}+t-1\end{array}$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The equation \(u^{\prime}(x)=\left(x^{2} u^{7}\right)^{-1}\) is separable. b. The general solution of the separable equation \(y^{\prime}(t)=\frac{t}{y^{7}+10 y^{4}}\) can be expressed explicitly with \(y\) in terms of \(t\) c. The general solution of the equation \(y y^{\prime}(x)=x e^{-y}\) can be found using integration by parts.

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y y^{\prime}(x)=\frac{2 x}{\left(2+y^{2}\right)^{2}}, y(1)=-1$$

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$\begin{array}{l} e^{-y / 2} y^{\prime}(x)=4 x \sin x^{2}-x ; y(0)=0 \\ y(0)=\ln \left(\frac{1}{4}\right), y(\sqrt{\frac{\pi}{2}})=0 \end{array}$$

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