Knowing the step size means we know each value of *x* we'll need to look at, so we can put these values in the table: *x* | *y*_{old} | slope × Δ *x* | *y*_{new} = *y*_{old} + slope × Δ *x* | 0 | | | | 0.4 | | | | 0.4 | | | | 0.6 | | | | 0.8 | | | | 1 | | | |
We know the initial value *y*(0) = 0, so we can put that in: Now we start the tangent line approximations. From the point (0,0) we calculate so we can fill in more of the table: Now we take *y*_{new} and use it as the value of *y*_{old} for the next step: When *x* = 0.2 and *y* = 0 the slope is 2(0.2) = .4, so we can fill in more: When *x* = 0.4 and *y* = 0.08 we get the next step: When *x* = .6 and *y* = .24 we get this: And finally, when *x* = .8 and *y* = .48 we get this: Since we now have a *y*-value corresponding to *x* = 1, we're done. We conclude that for the given IVP, Euler's method with Δ *x* = .2 estimates *y*(1) ≅ .8 An Euler's Method problem must include - an initial value
- enough information so you can figure out Δ
*x* - the stopping point (that is, the value of
*x* at which you want to approximate the function).
The previous example said "Let *y* be the solution to the differential equation that passes through the point (0,0). Use Euler's Method to estimate *y*(1) with step size Δ *x* = 0.2." The initial value is the point (0,0). Δ *x* = 0.2 was explicitly given. Since the problem says to estimate *y*(1), we know *x* = 1 is the stopping point. The problem could also have said "Use Euler's Method with 5 steps to estimate *y*(1)." In this case, the problem doesn't explicitly tell you Δ *x*. But it does tell you that you're supposed to get from *x* = 0 to *x* = 1 in 5 steps, which means |