Question

Carry out one step of the Euler method and of the improved Euler method using the step size \(h=0.1 .\) Suppose that a local truncation error no greater than 0.0025 is required. Estimate the step size that is needed for the Euler method to satisfy this requirement at the first step. $$ y^{\prime}=\left(y^{2}+2 t y\right) /\left(3+t^{2}\right), \quad y(0)=0.5 $$

Short answer

Using the Euler method, the first step approximation for the given differential equation is approximately y₁ ≈ 0.5083. For the Improved Euler method, the first step approximation is y₁ ≈ 0.5084. To achieve a local truncation error no greater than 0.0025, the step size required for the Euler method is approximately h ≈ 0.023.

Step-by-step solution

(1) Euler Method - First Step

Start with the differential equation \(y'(t) = \frac{y^2 + 2ty}{3+t^2}\), and apply the Euler method: $$y_{n+1} = y_n + hf(t_n, y_n),$$ where h = 0.1, \(f(t_n, y_n)\) is the value of the differential equation at the point \((t_n, y_n)\), and \((t_0, y_0) = (0, 0.5)\). First, calculate the value of the differential equation at the initial point, \(f(t_0, y_0) = f(0, 0.5) = \frac{(0.5)^2 + 2(0)(0.5)}{3 + (0)^2} = \frac{0.25}{3}\). Next, apply the Euler method with h = 0.1: $$y_1 = y_0 + 0.1\cdot f(t_0, y_0) = 0.5 + 0.1\cdot \frac{0.25}{3} \approx 0.5083.$$

(2) Improved Euler Method - First Step

Now, apply the Improved Euler method (also known as Heun's method). The formula for the improved Euler method is: $$y_{n+1} = y_n + \frac{h}{2}\left( f(t_n, y_n) + f(t_{n+1}, y_n+hf(t_n, y_n)) \right).$$ Use the previous result, \(f(t_0, y_0) = \frac{0.25}{3}\), and calculate \(f(t_1, y_1) = f(0.1, 0.5083) \approx \frac{(0.5083)^2 + 2(0.1)(0.5083)}{3 + (0.1)^2}\). Finally, apply the Improved Euler Method: $$y_1 \approx 0.5 + \frac{0.1}{2}\left( \frac{0.25}{3} \right) \approx 0.5084$$.

(3) Determine step size for required local truncation error

A local truncation error no greater than 0.0025 is required. The local truncation error for the Euler method can be estimated as: $$\text{Local Truncation Error} \approx \frac{hL}{2}|y(t_1) - y_0|,$$ where L is the Lipschitz constant. First, we need to find L. To do so, compute the partial derivative of the differential equation with respect to y: $$\frac{\partial f}{\partial y} = \frac{2y (3 + t^2) - 2yt (y^2 + 2ty)}{(3+t^2)^2}$$. Notice that the maximum value of the denominator occurs when t = 0, that is: $$(3+t^2)^2 \geq 3^2 = 9$$. Now, we estimate the maximum value of the numerator. Since \(y_0 = 0.5\), \(t=0\), we can find it as follows: $$\max_{0\leq y \leq y_1} |2y (3 + t^2) - 2yt (y^2 + 2ty)| \leq |2y_1 (3 + 0) - 2y_1^2|$$. So, we can estimate L as: $$L \leq \frac{|2y_1 (3) - 2y_1^2|}{9}$$. Now, plug L into the formula for the local truncation error and solve for h: $$0.0025 \geq \frac{hL}{2}|y_1 - y_0| \Rightarrow h \leq \frac{0.005}{L|y_1-y_0|}$$. Use the previous result and plug in the estimated L to find the appropriate step size h: $$h \leq \frac{0.005}{\frac{|2y_1 (3) - 2y_1^2|}{9}|y_1-y_0|} \approx 0.023$$. As such, the step size required to satisfy the local truncation error requirement is approximately h ≈ 0.023.