An initial-value problem is given by the differential equation,
f(x,y) = –20xy2, y(1) = 1.
Use the classical fourth-order Runge-Kutta method with a step-size of h = 0.02, to obtain the approximate value of y(1.02). Give your answer to 6 decimal places.
To solve the initial-value problem using the classical fourth-order Runge-Kutta method, we need to iterate through the steps of the method to approximate the value of y at the desired time point.
First, let's define the necessary variables:
- x0 = 1 (initial x value)
- y0 = 1 (initial y value)
- h = 0.02 (step-size)
- xf = 1.02 (final x value)
Now, let's go through the steps of the fourth-order Runge-Kutta method:
Step 1:
Calculate k1 = hf(x0, y0).
In this case, f(x0, y0) = -20xy^2, so k1 = 0.02 * (-20 * 1 * 1^2) = -0.4.
Step 2:
Calculate k2 = hf(x0 + h/2, y0 + k1/2).
Using the given values, k2 = 0.02 * (-20 * (1 + 0.02/2) * (1 + (-0.4)/2)) = -0.39992.
Step 3:
Calculate k3 = hf(x0 + h/2, y0 + k2/2).
Again, using the given values, k3 = 0.02 * (-20 * (1 + 0.02/2) * (1 + (-0.39992)/2)) = -0.39992.
Step 4:
Calculate k4 = hf(x0 + h, y0 + k3).
Once again, using the given values, k4 = 0.02 * (-20 * (1 + 0.02) * (1 + (-0.39992))) = -0.39984.
Step 5:
Calculate the approximate value of y at xf using the formula:
y(xf) = y0 + (k1 + 2k2 + 2k3 + k4)/6.
Plugging in the values, y(1.02) = 1 + (-0.4 + 2 * (-0.39992) + 2 * (-0.39992) + (-0.39984))/6 = 0.961106.
Therefore, the approximate value of y(1.02) using the fourth-order Runge-Kutta method with a step-size of h = 0.02 is 0.961106 (rounded to 6 decimal places).