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I've been working on a problem for some time and I seem to be able to solve only half of it.

I wonder if anyone can help me with it.

I want to calculate the right asymptote of a function F(t), i.e. a line p = m t + b such that F(t) approaches it for t[itex]\rightarrow \infty[/itex].

This would be pretty straightforward if I had an explicit definition p = F(t).

I don't.

What I have is a differential equation for another function G(t):

[itex]- \frac{dG(t)}{dt} = \frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V}[/itex]

where F(t) = Ln(G(t))

and a, Q, c and V are positive real constants.

The d.e. doesn't seem to be (easily) solvable.

So I tried another approach: leaving the function implicit.

The part that I solved by this approach was finding m (the slope of the asymptote).

I considered that:

[itex]- \frac{dF(t)}{dt} = - \frac{d Ln(G(t))}{dt} = - \frac{1}{G(t)}\frac{dG(t)}{dt} [/itex]

And note that from the theory that generated the above differential equation, I know that:

[itex]lim_{t \rightarrow \infty} G(t) = 0[/itex]

So I thought, as m is also F'(t) for t[itex]\rightarrow \infty[/itex], then:

[itex]m = lim_{t \rightarrow \infty} \frac{dF(t)}{dt} = lim_{t \rightarrow \infty} \frac{1}{G(t)}\frac{dG(t)}{dt} = - lim_{G(t) \rightarrow 0} \frac{a + Q G(t) + c - \sqrt{(a + Q G(t) + c)^{2} - 4 Q G(t) c}}{2 V G(t)}[/itex]

I don't know if this is mathematically 'legal', but I did it anyway. I submitted the expression to Maxima and this is what came out:

[itex]m = - \frac{Q c}{V (a + c)}[/itex]

which seems correct when I try it out graphically.

Now, the part that I don't know how to do is to find the general expression for the intercept q.

What I know from the theory is that:

[itex]q = lim_{t \rightarrow \infty} F(t) - m t[/itex]

but as I don't have F(t), I'm stuck.

I thought of substituting F(t) = Ln(G(t)), but that of course gives me -∞ +∞.

I wondered if I could use the Hopital rule, but I think that only applies to ratios.

So... any suggestions on how I could do this (or if it's at all possible in this case)?

Thanks!

L