First order ODE with $f'(x) = 810(10)^x$

Question

I'm trying to find an explicit form of the series $f(0) = 89.1,f(1) = 899.1,f(2) = 8999.1, \cdots$. My first though was to take the derivative and integrate it, which I've done before with a fair amount of success. I believe the derivative of this series is $810(10)^x$, and integrating we have $\frac{810(10)^x}{\ln10}+C$. solving for $C$, I get $89.1-\frac{810}{\ln10} \approx-262.679$, So the explicit form would be $\frac{810(10)^x}{\ln10}-262.679$. I know I went wrong, but I don't know where. If anyone could give me a general process to solve an ODE like this with an exponential as the derivative, and perhaps figure out how to solve this particular one, I would greatly appreciate it.

Answer 1

I would start by looking at $g(x) = f(x)+0.9$, then it takes values $90, 900, 9000, \ldots$ so it is simply a geometric series.

I think you measured the derivative incorrectly, it should be $$f'(x) = g'(x) = 10^x \cdot 90 \ln 10 \approx 207.23 \cdot 10^x$$

EDIT

Notice that if you define $g(x)$ as above then at $0,1,2, \ldots, g(x)$ evaluates to $90, 900, 9000, \ldots$, so $g(x) = 90 \cdot 10^x$ and thus $f(x) = g(x) - 0.9 = 90 \cdot 10^x - 0.9$.