To approximate the value of f(6) using Euler's Method, we will use the following steps:
Step 1: Determine the differential equation dy/dx = y - x
Step 2: Set up the initial condition y(5) = 1, which means that the point (5,1) is on the graph of the solution.
Step 3: Choose a step size, ∆x = 0.5, which represents the change in x as we move forward.
Step 4: Use Euler's Method to find the approximations for f(6) using the following iteration:
- Start with the initial condition: (x0, y0) = (5, 1)
- Compute the slope: m = y - x = f(x) - x
- Calculate the change in y: ∆y = m * ∆x = (f(x) - x) * ∆x
- Calculate the next y-value: y1 = y0 + ∆y
Step 5: Repeat the iteration for each step until we reach the desired value of x, which is x = 6 in this case.
Let's go through the steps to find the approximation of f(6) in more detail:
Step 1: The given differential equation is dy/dx = y - x.
Step 2: The initial condition is y(5) = 1, which means the starting point is (5, 1).
Step 3: The chosen step size is ∆x = 0.5.
Step 4: We will use Euler's Method to calculate the successive approximations for f(6) as follows:
Iteration 1:
- Start with the initial condition: (x0, y0) = (5, 1)
- Compute the slope: m = y - x = f(x) - x = y0 - x0 = 1 - 5 = -4
- Calculate the change in y: ∆y = m * ∆x = -4 * 0.5 = -2
- Calculate the next y-value: y1 = y0 + ∆y = 1 - 2 = -1
Iteration 2:
- Start with the previous point: (x0, y0) = (5, -1)
- Compute the slope: m = y - x = f(x) - x
- In this case, we have to calculate f(x) at x = 5 because we don't have an explicit equation for f(x).
- Using the given differential equation, dy/dx = y - x, we can substitute y = f(x) into the equation: df(x)/dx = f(x) - x.
- Therefore, at x = 5, dy/dx = f(5) - 5.
- Calculate the change in y: ∆y = m * ∆x = (f(5) - 5) * 0.5 = (y0 - x0) * ∆x = (-1 - 5) * 0.5 = -6
- Calculate the next y-value: y1 = y0 + ∆y = -1 - 6 = -7
Step 5: We continue this process, repeating the iterations until we achieve the desired x-value, which is x = 6.
In summary, using Euler's Method with a step size of ∆x = 0.5, we find that the approximation of f(6) is -7.