Let 𝑦=𝑓(𝑥) be the solution to the differential equation 𝑑𝑦𝑑𝑥=2𝑥+𝑦 with initial condition 𝑓(1)=0. What is the approximation for 𝑓(2) obtained by using Euler’s method with two steps of equal length, starting at
Question
Let 𝑦=𝑓(𝑥) be the solution to the differential equation 𝑑𝑦𝑑𝑥=2𝑥+𝑦 with initial condition 𝑓(1)=0. What is the approximation for 𝑓(2) obtained by using Euler’s method with two steps of equal length, starting at
Solution
To solve this problem using Euler's method, we first need to understand that Euler's method is a simple way to approximate solutions to first-order differential equations. The general formula for Euler's method is:
y_(n+1) = y_n + h*f(x_n, y_n)
where h is the step size, f(x_n, y_n) is the derivative at the point (x_n, y_n), and y_(n+1) is the next approximation of the solution.
Given the differential equation dy/dx = 2x + y and the initial condition f(1) = 0, we can start applying Euler's method.
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First Step:
- Our starting point is (1, 0).
- The derivative at this point is f'(1, 0) = 2*1 + 0 = 2.
- We are told to use two steps of equal length, so from x = 1 to x = 2 with two steps means each step is of size 0.5.
- Applying Euler's method, we get the next y as: y_(n+1) = y_n + hf(x_n, y_n) = 0 + 0.52 = 1.
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Second Step:
- Our new starting point is (1.5, 1).
- The derivative at this point is f'(1.5, 1) = 2*1.5 + 1 = 4.
- Again, our step size is 0.5.
- Applying Euler's method, we get the next y as: y_(n+1) = y_n + hf(x_n, y_n) = 1 + 0.54 = 3.
So, the approximation for f(2) obtained by using Euler’s method with two steps of equal length, starting at x = 1, is 3.
Similar Questions
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