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Use Euler's method with step size 0.5 to compute the approximate y-values y1, y2, y3 and y4 of the solution of the initial-value problem y' = y − 5x, y(3) = 0.y1 = y2 = y3 = y4 =

Question

Use Euler's method with step size 0.5 to compute the approximate y-values y1, y2, y3 and y4 of the solution of the initial-value problem y' = y − 5x, y(3) = 0.y1 = y2 = y3 = y4 =

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Solution

To use Euler's method with a step size of 0.5, we need to follow these steps:

Step 1: Determine the number of iterations. In this case, we need to compute y1, y2, y3, and y4, so we will have four iterations.

Step 2: Set up the initial conditions. The initial condition given is y(3) = 0. Therefore, we start with x0 = 3 and y0 = 0.

Step 3: Calculate the approximate y-values using Euler's method. The formula for Euler's method is:

yn+1 = yn + h * f(xn, yn)

where yn+1 is the next y-value, yn is the current y-value, xn is the current x-value, h is the step size, and f(xn, yn) is the derivative of y with respect to x evaluated at xn and yn.

Using the given initial-value problem, we have f(x, y) = y - 5x.

For the first iteration (n = 1): x1 = x0 + h = 3 + 0.5 = 3.5 y1 = y0 + h * f(x0, y0) = 0 + 0.5 * (0 - 5 * 3) = -7.5

For the second iteration (n = 2): x2 = x1 + h = 3.5 + 0.5 = 4 y2 = y1 + h * f(x1, y1) = -7.5 + 0.5 * (-7.5 - 5 * 3.5) = -16.25

For the third iteration (n = 3): x3 = x2 + h = 4 + 0.5 = 4.5 y3 = y2 + h * f(x2, y2) = -16.25 + 0.5 * (-16.25 - 5 * 4) = -27.125

For the fourth iteration (n = 4): x4 = x3 + h = 4.5 + 0.5 = 5 y4 = y3 + h * f(x3, y3) = -27.125 + 0.5 * (-27.125 - 5 * 4.5) = -40.5625

Therefore, the approximate y-values are: y1 = -7.5 y2 = -16.25 y3 = -27.125 y4 = -40.5625

This problem has been solved

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