*Differential Equations With Boundary Value Problems*by Polking, Boggess, and Arnold...

Exercise 12 references Exercise 11, to which the official solution is...

Exercise 12 part (a) is straightforward; solve [imath]\frac{dQ}{dt} = .06Q,\ Q(0) = 2000, Q(10) =[/imath] the solution, which turns out as [imath]\$3644.24[/imath].

I think the point of this problem requested by part (c) is that the solutions to the

*difference*equations at the [imath]10[/imath] year point are supposed to monotonically increase as we increase [imath]m[/imath], and converge to the

*differential*equation solution as [imath]m[/imath] approaches [imath]\infty[/imath]. Since there's nothing special about the [imath]10[/imath] year point, we can be slightly more ambitious and expect the entire particular solution curve to converge as such.

I'm just a bit stuck on how to get there by following part (b). The formula generated in Exercise 11, [imath]P(n) = 2000(1 + \frac{.06}{m})^n[/imath], doesn't converge to the differential equation solution [imath]Q(t) = 2000e^{.06t}[/imath] for large [imath]m[/imath], but [imath]P[n] = 2000(1 + \frac{.06}{m})^{mn}[/imath] does, so I think I need to get an extra [imath]m[/imath] into the exponent.

I could apply the substitution [imath]t = \frac{n}{m}[/imath], [imath]Q(t) = P(n)[/imath] to the difference equation solution to get it into the right form, but this feels superficial; the substitution would never be undone, and I could just as easily make any substitution to get almost any desired answer.

What else can I do to introduce this missing [imath]m[/imath] in a careful fashion?