Use Euler's method to find a numerical solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}=-2 y, y(0)=1\), for \(0 \leq x \leq 0.5 .\) First take \(h=0.1\), then \(h=0.05\), and compare your answers at \(x=0.5\) with the exact solution obtained by separating the variables. Work throughout to six decimal places.

Short Answer

Expert verified
Using Euler's method, we calculated the approximate value of \(y(0.5)\) for two different step sizes, \(h=0.1\) and \(h=0.05\). For \(h=0.1\), the approximate value of \(y(0.5)\) is 2.488, while for \(h=0.05\), it is 2.362. In comparison with the exact solution \(y(0.5) \approx 0.3679\), neither of the estimated values achieved good accuracy. The solution with \(h=0.05\) was slightly closer to the exact solution than the one with \(h=0.1\), but to obtain better accuracy, it is necessary to either increase the number of steps or use another numerical method.

Step by step solution

01

Review Euler's Method

Euler's method estimates the value of the unknown function as it moves along the independent variable for small intervals. The general formula to apply Euler's method is: \(y_{n+1} = y_n + h*f(x_n, y_n)\), where \(h\) is the step size, \(y_n\) is the current value of \(y\), \(x_n\) is the current value of \(x\), and \(f(x_n, y_n)\) is the function representing the derivative of \(y\) with respect to \(x\). In this problem, \(f(x, y) = -2y\).
02

Apply the Euler's method for \(h=0.1\)

Let's iterate the Euler's method with \(h = 0.1\), starting from \(x_0=0\) and \(y_0= 1\): 1. For \(n=0\): \(f(x_0, y_0) = -2(1) = -2\). Hence, \(y_1 = y_0 + h * f(x_0, y_0) = 1 -0.1*(-2) = 1+0.2=1.2\) 2. For \(n=1\): \(f(x_1, y_1) = -2(1.2) = -2.4\). Hence, \(y_2 = y_1 + h * f(x_1, y_1) = 1.2 -0.1*(-2.4) = 1.2+0.24=1.44\) 3. For \(n=2\): \(f(x_2, y_2) = -2(1.44) = -2.88\). Hence, \(y_3 = y_2 + h * f(x_2, y_2) = 1.44 -0.1*(-2.88) = 1.44+0.288=1.728\) 4. For \(n=3\): \(f(x_3, y_3) = -2(1.728) = -3.456\). Hence, \(y_4 = y_3 + h * f(x_3, y_3) = 1.728 -0.1*(-3.456) = 1.728+0.345=2.073\) 5. For \(n=4\): \(f(x_4, y_4) = -2(2.073) = -4.146\). Hence, \(y_5 = y_4 + h * f(x_4, y_4) = 2.073 -0.1*(-4.146) = 2.073+0.415 =2.488\) So, we got \(y_5 \approx 2.488\), which corresponds to the value of \(y(0.5)\).
03

Apply the Euler's method for \(h=0.05\)

Iterating the Euler's method with \(h = 0.05\), increasing the number of steps, and repeating the process: For \(n=0\): \(y_1 \approx 1.100\) For \(n=1\): \(y_2 \approx 1.210\) For \(n=2\): \(y_3 \approx 1.331\) For \(n=3\): \(y_4 \approx 1.464\) For \(n=4\): \(y_5 \approx 1.611\) For the rest of the steps, the results are as follows: For \(n=6\): \(y_7 \approx 1.774\) For \(n=7\): \(y_8 \approx 1.952\) For \(n=8\): \(y_9 \approx 2.147\) For \(n=9\): \(y_{10} \approx 2.362\) So, we got \(y_{10} \approx 2.362\), at the end.
04

Calculate exact solution and compare results

In order to obtain the exact solution, we need to first understand the differential equation by separating variables: \(\frac{dy}{dx} = -2y\) \(\frac{dy}{y} = -2 dx\) Integrating both sides and using the initial condition \(y(0)=1\), we get \(\ln|y| = -2x + C\) As \(y(0) = 1\), this implies that \(C = 0\). Therefore, the exact solution is \(y(x) = e^{-2x}\) Now, let's calculate \(y(0.5)\) using the exact solution: \(y(0.5) = e^{-2(0.5)} = e^{-1} \approx 0.3679\) Now, let's compare the estimated results for \(h=0.1\) and \(h=0.05\) with the exact solution: \(h=0.1\) -> \(y(0.5) \approx 2.488\) \(h=0.05\) -> \(y(0.5) \approx 2.362\) Exact solution -> \(y(0.5) \approx 0.3679\) From the comparison, we can see that the approximate value for \(y(0.5)\) using Euler's method is not very accurate, and the solution with \(h=0.05\) is closer to the exact solution than \(h=0.1\). It is necessary to either increase the step amount or use another numerical method to achieve better accuracy.

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