Question

Euler's method:
Let y=f(x) be the solution to the differential equation dy/dx = arcsin(xy) with the initial condition f(0)=2. What is the approximation for f(1) if Euler's method is used, starting at x=0 with a step size of .5?

Answer 1

Answer:

f(1) ≈ 2.7864

Step-by-step explanation:

You appear to want a couple of iterations of ...

... y[n+1] = y[n] +arcsin(x[n]·y[n]}·(x[n+1] -x[n])

... x[n+1] = x[n] +0.5

... x[0] = 0

... y[0] = 2

Filling in the values, we get

... y[1] = 2 + arcsin(0·2)·0.5 = 2

... y[2] = 2 + arcsin(0.5·2)·0.5 = 2 +(π/2)·0.5 ≈ 2.7864 . . . . corresponds to x=1