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Find y(2.2) using Euler's method from equation dy dx = -xy² and y(2)=1 for 4 intervals.?
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Find y(2.2) using Euler's method from equation dy dx = -xy² and y(2)=1...
To find y(2.2) using Euler's method from the given equation dy/dx = -xy² and y(2)=1 for 4 intervals, we can follow the below steps:

Euler's Method:

Euler's method is a numerical method to solve ordinary differential equations (ODEs). It is a simple and straightforward method that is based on the idea of approximating the solution at a given point using the derivative of the function at that point.

Steps:

1. Define the problem: We are given the ODE dy/dx = -xy² and the initial condition y(2) = 1. We need to find the value of y(2.2) using Euler's method with 4 intervals.

2. Determine the step size: The step size h is the distance between two consecutive points on the x-axis. In this case, we need to divide the interval [2, 2.2] into 4 equal subintervals. So, the step size is h = (2.2 - 2)/4 = 0.05.

3. Set up the Euler's method formula: Euler's method formula is given by y1 = y0 + hf(x0, y0), where y1 is the approximation of y at the next point, y0 is the known value of y at the current point, h is the step size, x0 is the current point on the x-axis, and f(x0, y0) is the derivative of the function at (x0, y0).

4. Calculate the approximation of y at each point: We can use the formula from step 3 to calculate the approximation of y at each point. We start with y0 = 1 and x0 = 2, and then use the formula to calculate y1, y2, y3, and y4.

5. Calculate y(2.2): Once we have calculated the approximations of y at each point, we can use the value of y at the last point (y4) as the approximation of y(2.2).

Calculation:

1. Define the problem: dy/dx = -xy², y(2) = 1, and we need to find y(2.2) using Euler's method with 4 intervals.

2. Determine the step size: h = (2.2 - 2)/4 = 0.05.

3. Set up the Euler's method formula: y1 = y0 + hf(x0, y0), where f(x, y) = -xy².

4. Calculate the approximation of y at each point:

- At x0 = 2, y0 = 1.
- At x1 = 2.05, y1 = y0 + h*(-x0*y0²) = 1 + 0.05*(-2*1²) = 0.9.
- At x2 = 2.1, y2 = y1 + h*(-x1*y1²) = 0.9 + 0.05*(-2.05*0.9²) = 0.818.
- At x3 = 2.15, y3 = y2 + h*(-x2*y2²) = 0.818 + 0.05*(-2.1*0.818²) = 0.756
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