1.10E: Exercises for Section 1.10

Exercise \(\PageIndex{1}\)

Consider \(\frac{dx}{dt} = (2t - x)^2\), \(x(0) = 2\). Use Euler’s method with step size \(h = 0.5\) to approximate \(x(1)\).

Answer

\( x(1) \approx 8.5 \)

Exercise \(\PageIndex{2}\)

Consider \(\frac{dx}{dt} = t -x \), \(x(0) = 1\).

1. Use Euler’s method with step sizes \(h = 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}\) to approximate \(x(1)\).
2. Solve the equation exactly.
3. Describe what happens to the errors for each \(h\) you used. That is, find the factor by which the error changed each time you halved the interval.

Hint: 

Euler update formula: \( x_{n+1} = x_n + h(t_n - x_n) \)

Runge–Kutta Method (RK4) formula: \( 
k_1 = f(x_n, y_n), \quad k_2 = f(x_n + h/2, y_n + hk_1/2),\\
k_3 = f(x_n + h/2, y_n + hk_2/2), \quad k_4 = f(x_n + h, y_n + hk_3)\\
y_{n+1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \)

Answer

1.  Approximations of  \(x(1)\):
   \( \begin{array}{c|c}
   h & x(1)\ \text{(Euler Approximation)} \\ \hline
   1 & 0 \\
   \frac{1}{2} & 0.5 \\
   \frac{1}{4} & 0.578125 \\
   \frac{1}{8} & 0.6478
   \end{array}
   \)
2. \( x(1) = \frac{2}{e} \approx 0.73576 \)
3. \( \begin{array}{c|c|c}
   h & \text{Approximation} & \text{Error} & \textbf{ Error Reduction}\\ \hline
   1 & 0 & 0.73576 \\
   \frac{1}{2} & 0.5 & 0.23576 & \frac{E_{1/2}}{E_1} \approx 0.32\\
   \frac{1}{4} & 0.578125 & 0.15763 & \frac{E_{1/4}}{E_{1/2}} \approx 0.67 \\
   \frac{1}{8} & 0.6478 & 0.08796 & \frac{E_{1/8}}{E_{1/4}} \approx 0.56
   \end{array} \) 

Exercise \(\PageIndex{3}\)

Approximate the value of \(e\) by looking at the initial value problem \(y' = y\) with \(y(0) = 1\) and approximating \(y(1)\) using Euler’s method with a step size of \(0.2\).

Answer

 \(  e \approx 2.48832  \)

Exercise \(\PageIndex{4}\)

The simplest method used in practice is the Runge-Kutta method. Consider \(\frac{dy}{dx} = f(x, y)\), \(y(x_0) = y_0\) and a step size \(h\). Everything is the same as in Euler’s method, except the computation of \(y_{i+1}\) and \(x_{i+1}\).

\[\begin{align}\begin{aligned} k_1 &= f(x_i, y_i), \\ k_2 &= f(x_i + \frac{h}{2}, y_i +k_1 \frac{h}{2}) & x_{i+1} = x_i + h, \\ k_3 &= f(x_i + \frac{h}{2}, y_i +k_2 \frac{h}{2}) & y_{i+1} = y_i + \dfrac{k_1 + 2k_2 + 2k_3 + k_4}{6}h, \\ k_4 &= f(x_i + h, y_i +k_3h).\end{aligned}\end{align}\]

Exercise \(\PageIndex{5}\)

Consider \(\frac{dy}{dx} = yx^2\), \(y(0) = 1\).

1. Use Runge-Kutta (see Exercise 1.10E.4) with step sizes \(h = 1\) and \(h = \frac{1}{2}\) to approximate \(y(1)\).
2. Use Euler’s method with \(h = 1\) and \(h = \frac{1}{2}\).
3. Solve exactly, find the exact value of \(y(1)\), and compare.

Answer

1. \(  h = 1: y(1) \approx 1.3906,  \), \(  h = \frac{1}{2}: y(1) \approx 1.3967 \)
2. \( y(1) \approx 1, y(1) \approx 1.125 \)
3. \( y(1) = e^{1/3} \approx 1.3956 \)

Exercise \(\PageIndex{6}\)

Let \(x' = \sin (xt)\), and \(x(0) = 1\). Approximate \(x(1)\) using Euler’s method with step sizes \(1\), \(0.5\), \(0.25\). Use a calculator and compute up to \(4\) decimal digits.

Answer

Approximately: \(1.0000,\: 1.2397,\: 1.382\)

Exercise \(\PageIndex{7}\)

Let \(x' = 2t\), and \(x(0) = 0\).

1. Approximate \(x(4)\) using Euler’s method with step sizes \(4\), \(2\), and \(1\).
2. Solve exactly, and compute the errors.
3. Compute the factor by which the errors changed.

Answer

1. \(0,\: 8,\: 12\)
2. \(x(4) = 16,\) so errors are: \(16,\: 8,\: 4\).
3. Factors are \(0.5,\: 0.5,\: 0.5\).

Exercise \(\PageIndex{8}\)

Let \(x' = xe^{xt+1}\), and \(x(0) = 0\).

1. Approximate \(x(4)\) using Euler’s method with step sizes \(4\), \(2\), and \(1\).
2. Guess an exact solution based on part (a) and compute the errors.

Answer

1. \(0,\: 0,\: 0\)
2. \(x=0\) is a solution so errors are: \(0,\: 0,\: 0\).

Exercise \(\PageIndex{9}\)

Consider \(\dfrac{dy}{dx} = x+y\), \(y(0)=1\).

1. Use the improved Euler’s method with step sizes \(h=\frac{1}{4}\) and \(h=\frac{1}{8}\) to approximate \(y(1)\).
2. Use Euler’s method with \(h=\frac{1}{4}\) and \(h=\frac{1}{8}\).
3. Solve exactly, find the exact value of \(y(1)\).
4. Compute the errors, and the factors by which the errors changed.

Answer

1. Improved Euler: \(y(1)\approx 3.3897\) for \(h=1/4\), \(y(1)\approx 3.4237\) for \(h=1/8\),
2. Standard Euler: \(y(1)\approx 2.8828\) for \(h=1/4\), \(y(1)\approx 3.1316\) for \(h=1/8\),
3. \(y=2e^{x}-x-1\), so \(y(2)\) is approximately \(3.4366\).
4. Approximate errors for improved Euler: \(0.046852\) for \(h=1/4\), and \(0.012881\) for \(h=1/8\). For standard Euler: \(0.55375\) for \(h=1/4\), and \(0.30499\) for \(h=1/8\). Factor is approximately \(0.27\) for improved Euler, and \(0.55\) for standard Euler.