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Obtain a formula for the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\) $$ y^{\prime}=5 t-3 \sqrt{y}, \quad y(0)=2 $$

Short Answer

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Answer: Without an explicit form for the exact solution, we can't obtain an exact expression for the local truncation error. However, for any given exact solution \(\phi(t)\) of the differential equation, the truncation error is represented in terms of \(t\) and the solution \(\phi\) as: $$E = \phi(t_1) - y_1 = \phi(t_0 + h) - (2 - 3h\sqrt{2})$$ This expression shows the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\) for the given differential equation.

Step by step solution

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01

Define the Euler method and the given differential equation

We have the following differential equation: $$y^{\prime}=5 t-3 \sqrt{y}, \quad y(0)=2$$ The Euler method is a numerical method to approximate the solution to this type of differential equation, and it is defined as follows: $$ y_{n+1} = y_n + hf(t_n, y_n)$$ where \(h\) is the step size and \(f(t, y)\) is the function representing the derivative.
02

Take the derivative of the exact solution, \(\phi'(t)\)

We don't have an explicit form for the exact solution \(\phi(t)\) of the differential equation. However, we know that it satisfies the differential equation, so its derivative \(\phi^{\prime}(t)\) can be expressed as: $$ \phi^{\prime}(t) = 5t - 3 \sqrt{\phi(t)} $$
03

Calculate the error

To find the local truncation error, we need to compare the Euler method's approximation with the exact solution. Using the Euler method, the estimate after the first step is: $$ y_1 = y_0 + h \cdot f(t_0, y_0) $$ Substitute the given derivative \(y^{\prime}\) and initial conditions \(y(0) = 2\), \(t_0 = 0\), we have: $$ y_1 = 2 + h(5(0) - 3\sqrt{2}) = 2 - 3h\sqrt{2} $$ The error \(E\) after the first step is the difference between the exact solution at \(t_1 = t_0 + h\) and the Euler method's approximation: $$E = \phi(t_1) - y_1 $$ Notice that \(E\) depends on \(t_1\) and \(\phi(t_1)\).
04

Substitute the exact solution and its derivative into the error formula

Unfortunately, we can't obtain an explicit error formula without knowing the exact solution \(\phi(t)\). But, for any given exact solution of the differential equation, the truncation error is expressed in terms of \(t\) and the solution \(\phi\) as: $$ E = \phi(t_1) - y_1 = \phi(t_0 + h) -(2 - 3h\sqrt{2})$$ This expression represents the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\) for the given differential equation.

Key Concepts

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

Local Truncation Error
When employing numerical methods like the Euler Method, one key aspect to consider is the local truncation error (LTE). This error refers to the difference incurred at each step between the true solution and the approximation obtained using the method. Essentially, it assesses how well the numerical method approximates the differential equation at a single step rather than across an entire interval.

For the Euler Method, the local truncation error can be seen as the discrepancy created from using a linear approximation. Typically, the Euler approximation involves a step-wise calculation, and the LTE arises from approximating the tangent line rather than the curve of the actual function.

  • The LTE depends on both the function being evaluated and the step size \( h \).
  • Smaller step sizes generally lead to smaller local truncation errors, promoting accuracy.
  • In the context of the given exercise, the error is formulated as \( E = \phi(t_1) - y_1 = \phi(t_0 + h) -(2 - 3h\sqrt{2}) \).
Understanding LTE helps in selecting an appropriate step size and assessing the reliability of the numerical approximation.
Differential Equation
At the core of many mathematical problems is the concept of a differential equation, which represents a relationship involving derivatives and functions. Specifically, a differential equation often seeks to define a function based on its rate of change.

In this exercise, we are dealing with a first-order differential equation given by \( y^{\prime}=5t-3\sqrt{y} \), along with an initial condition \( y(0)=2 \). This means the derivative of the function \( y \) is expressed in terms of the variable \( t \) and the function \( y \) itself.

  • Differential equations can be linear or nonlinear; the given equation is nonlinear due to the square root term.
  • They are used to model various real-world phenomena such as population growth, physics problems, or even economics.
  • Solving a differential equation involves finding a function or set of functions that satisfies the equation.
In practice, finding exact solutions analytically can be intricate, hence the application of numerical methods like Euler's Method.
Numerical Methods
Numerical methods are techniques designed to approximate solutions to mathematical problems where analytical solutions are difficult or impossible to obtain directly. These can handle complex differential equations, systems of equations, and other computational tasks.

The Euler Method, showcased in this exercise, is a fundamental numerical technique. It approximates the solution to an initial value problem by generating successive approximations from an initial point.

  • These methods involve calculations performed in steps, each portrayed by a specific size \( h \), known as the step size.
  • They're employed in cases where only derivative information is available, as in this exercise.
  • Other common numerical methods include the Runge-Kutta methods, which offer improved accuracy over Euler's method by considering intermediate points.
Choosing the right numerical method often balances accuracy with computational efficiency and can vary based on the specific problem requirements.
Initial Value Problem
An initial value problem (IVP) involves solving a differential equation with a given initial condition. This means the solution is subject to a specific starting point, allowing it to be uniquely defined.

In the given exercise, we tackle an IVP with the initial condition \( y(0)=2 \), where the function \( y \) is defined at \( t=0 \). Initial value problems are essential in many scientific and engineering fields because they reflect scenarios where a process begins from a known state.

  • IVPs are solved through integrating the differential equation while respecting the initial conditions.
  • This approach helps ensure the solution curve passes through the known point \( y(0)=2 \) at \( t=0 \).
  • In practical terms, they model situations like starting at a specific time or initial population size.
Utilizing numerical methods is often practical for solving IVPs, especially when analytical solutions are hard to derive due to complexity. Euler's Method, as used here, facilitates handling these problems, albeit with some approximation.

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Most popular questions from this chapter

The modified Euler formula for the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) is given by $$ y_{n+1}=y_{n}+h f\left[t_{n}+\frac{1}{2} h, y_{n}+\frac{1}{2} h f\left(t_{n}, y_{n}\right)\right] $$ Following the procedure outlined in Problem \(14,\) show that the local truncation error in the modified Euler formula is proportional to \(h^{3}\).

Consider the initial value problem $$ y^{\prime}=t^{2}+e^{y}, \quad y(0)=0 $$ Using the Runge-Kutta method with step size \(h,\) we obtain the results in Table \(8.5 .\) These results suggest that the solution has a vertical asymptote between \(t=0.9\) and \(t=1.0 .\) (a) Show that for \(0 \leq t \leq 1\) the solution \(y=\phi(t)\) of the problem (i) satisfies $$ \phi_{2}(t) \leq \phi(t) \leq \phi_{1}(t) $$ where \(y=\phi_{1}(t)\) is the solution of $$ y^{\prime}=1+e^{y}, \quad y(0)=0 $$ and \(y=\phi_{2}(t)\) is the solution of $$ y^{\prime}=e^{y}, \quad y(0)=0 $$ (b) Determine \(\phi_{1}(t)\) and \(\phi_{2}(t) .\) Then show that \(\phi(t) \rightarrow \infty\) for some \(t\) between \(t=\ln 2 \cong\) 0.69315 and \(t=1 .\) (c) Solve the differential equations \(y^{\prime}=e^{y}\) and \(y^{\prime}=1+e^{y},\) respectively, with the initial condition \(y(0.9)=3.4298 .\) Use the results to show that \(\phi(t) \rightarrow \infty\) when \(t \cong 0.932 .\)

Consider the initial value problem $$ y^{\prime}=t^{2}+y^{2}, \quad y(0)=1 $$ (a) Draw a direction field for this equation. (b) Use the Runge-Kutta or other methods to find approximate values of the solution at \(t=0.8,0.9,\) and 0.95 . Choose a small enough step size so that you believe your results are accurate to at least four digits. (c) Try to extend the calculations in part (b) to obtain an accurate approximation to the solution at \(t=1\) If you encounter difficulties in doing this, explain why you think this happens. The direction field in part (a) may be helpful.

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 \(h=0.05\) $$ y^{\prime}=2 y-3 t, \quad y(0)=1 $$

(a) Find a formula for the solution of the initial value problem, and note that it is independent of \(\lambda\). (b) Use the Runge-Kuta method with \(h=0.01\) to compute approximate values of the solution for \(0 \leq t \leq 1\) for various values of \(\lambda\) such as \(\lambda=1,10,20,\) and 50 , and 50 , inters of the (c) Explain the differences, if any, between the exact solution and the numerical approximations. \(y^{\prime}-\lambda y=1-\lambda t, \quad y(0)=0\)

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