Problem 13 - Thomas Calculus

Question

In Exercises \(11-16,\) use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$ y^{\prime}=2 x y+2 y, \quad y(0)=3, \quad d x=0.2 $$

Step-by-Step Solution

Answer

Using Euler's method, the first three approximations are 4.2000, 6.2160, and 9.6970. The exact solution shows they are close but diverge as the calculations progress.

Step 1: Understanding Euler's Method

Euler's method is used to find numerical solutions to initial value problems. It starts with the initial condition and then proceeds in steps of size \(dx\), updating the current function value using the derivative formula \(y' = f(x, y)\). In this problem, we start with \(y_0 = 3\) at \(x_0 = 0\) and the increment size \(dx = 0.2\).

Step 2: Calculating the First Approximation

Using Euler's method, the formula to find the next value is \(y_{n+1} = y_n + dx \, f(x_n, y_n)\). With \(y' = 2xy + 2y\), calculate for \(x_0 = 0\) and \(y_0 = 3\):\(f(0, 3) = 2(0)(3) + 2(3) = 6\).Now, calculate \(y_1\):\(y_1 = 3 + 0.2 \times 6 = 3 + 1.2 = 4.2\).

Step 3: Calculating the Second Approximation

Proceed to calculate \(y_2\) for \(x_1 = 0.2\) and \(y_1 = 4.2\):\(f(0.2, 4.2) = 2(0.2)(4.2) + 2(4.2) = 1.68 + 8.4 = 10.08\).Now, calculate \(y_2\):\(y_2 = 4.2 + 0.2 \times 10.08 = 4.2 + 2.016 = 6.216\).

Step 4: Calculating the Third Approximation

Now calculate \(y_3\) for \(x_2 = 0.4\) and \(y_2 = 6.216\):\(f(0.4, 6.216) = 2(0.4)(6.216) + 2(6.216) = 4.9728 + 12.432 = 17.4048\).Calculate \(y_3\):\(y_3 = 6.216 + 0.2 \times 17.4048 = 6.216 + 3.48096 = 9.69696\).Rounded: \(y_3 = 9.6970\).

Step 5: Calculating the Exact Solution

The differential equation is separable. Write it as \(\frac{dy}{y} = (2x+2)dx\). Integrate both sides:\(\ln|y| = x^2 + 2x + C\).Using the initial condition \(y(0) = 3\), we find \(C\):\(\ln 3 = C\), so the exact solution is:\(y = 3e^{x^2 + 2x}\).

Step 6: Investigating Accuracy

Compute the exact values using the solution \(y = 3e^{x^2 + 2x}\) for \(x = 0.2, 0.4, 0.6\) and compare to approximations:- For \(x=0.2\), exact: \(y(0.2) = 3e^{0.44} \approx 4.2733\)- For \(x=0.4\), exact: \(y(0.4) = 3e^{1.36} \approx 6.2951\)- For \(x=0.6\), exact: Exact calculations neededCompare them to the approximated values for precision.

Key Concepts

Numerical Solutions

Euler's method is a simple yet powerful tool for deriving numerical solutions to differential equations, especially when an analytical solution is difficult or impossible to determine. This approach is particularly useful for initial value problems, where we know the starting conditions of a function and need to predict future values.
To apply Euler's method, we iteratively build upon initial conditions using small step increments. For each step, we use the formula:
\( y_{n+1} = y_n + dx \, f(x_n, y_n) \)
Here, \( f(x_n, y_n) \) is the derivative evaluated at \( (x_n, y_n) \), and \( dx \) is the chosen step size. This method essentially projects the tangent line of the curve forward over a small distance to approximate the next point on the curve.
In practice, smaller step sizes tend to yield more accurate approximations but require more computational work. The charm of numerical solutions is their ability to provide a scaffold for understanding and approximating functions without solving complex equations analytically.

Initial Value Problems

Initial value problems (IVPs) are a pivotal part of mathematical modeling, characterizing scenarios where the value of a solution to a differential equation is known at a specific point. In our example, we were given \( y(0) = 3 \) along with the differential equation \( y' = 2xy + 2y \).
In IVPs, the 'initial value' gives a starting point and is essential for defining the specific path the solution will take in the solution space. Without this initial value, the solutions could wildly vary since differential equations often have a family of solutions.

• The initial value ensures the function meets a certain condition at a specific point, making the problem well-posed.
• Well-posed problems are crucial as they guarantee that the problem has a solution, and that the solution behaves continuously with respect to initial data.

Applying a method like Euler's helps us explore the path of the solution as it evolves from this initial condition, providing practical insights into the behavior of dynamical systems.

Accuracy of Approximations

The accuracy of approximations is a key concern when utilizing numerical methods like Euler's. Given the stepwise nature of these techniques, there are always trade-offs between precision and computational efficiency. In our exercise, Euler's method provided approximate values for \( y \), but how close were these to the exact solution?
As we calculated in the exact solution step, the precise outcomes at points \( x = 0.2 \) and \( x = 0.4 \) were \( 4.2733 \) and \( 6.2951 \), respectively, while Euler's approximations gave us \( 4.2 \) and \( 6.216 \).

• The discrepancies indicate that while Euler's method offers a quick approximate solution, minor errors accumulate with each step.
• Reducing the step size \( dx \) would improve accuracy but at the cost of increasing the number of calculations needed.

Understanding this balance is pivotal for using numerical methods effectively. Whether solving through Euler's approach or using more sophisticated algorithms, the goal is to achieve a desirable accuracy without excessive computational expense.