In problems 1 - 6, let h = 0.1 and approximate the solutions to y(x_1) and y(x_2
ID: 2881763 • Letter: I
Question
In problems 1 - 6, let h = 0.1 and approximate the solutions to y(x_1) and y(x_2) by hand using (a) Euler's Method and (b) the fourth-order Runge-Kutta method. Then compare the results with the given explicit solution at x_1 and x_2 It will be easiest if you display the results as we did in the table of Figure 2.25. 1. y' = x^3, y(1) = 1; explicit solution: y = 1/4(x^4 + 3) 2. y' = -y^2, y(0) = 1; explicit solution: y = 1/x + 1 3. y' = x^4y, y(1) = 1; explicit solution: y = e^(x^5 - 1)/5 4. y' = -y^2 cos x, y(0) = 1; explicit solution: y = 1/1 + sin x 5. y' = sin x/y^3, y(pi) = 2; explicit solution: y = (12 - 4 cos x)^1/4 6. y' = ye^-x, y(0) = 1; explicit solution: y = exp(1 - e^-x)Explanation / Answer
Solution:
Euler's method, This is a numerical method for getting an approximate solution to the equation.
y ' (x) = x^3
use h = x = 0.1
we start with y_0 = y(x_0) = 1
y_1 = y_0 + x * y ' (x_0) = y_0 + 0.1 * (1)^3 = 1 + 0.1(1) = 1.1
y_2 = y_1 + x * y ' (x_1) = 1.1 + (0.1)(1.1)^3 = 1.1 + (0.1)(1.331) = 1.2331