Given that f(z) is a strictly increasing function

calcolare45pj

calcolare45pj

Answered question

2022-03-27

Given that f(z) is a strictly increasing function (with real input and output and no root) for which there exists some positive (real) value of b such that it asymptotically outgrows zb+1 for positive z.
Is it true that any solution to
dydx=f(y)
will contain a vertical asymptote providing y (real) exceeds 0 at some point?
Conversely, is such a vertical asymptote impossible given that zb+1 outgrows f(z)?

Answer & Explanation

Ashley Olson

Ashley Olson

Beginner2022-03-28Added 12 answers

Explanation:
The central computation is not that complicated. Set u=yb, then
u=byb1y=b·f(y)y1+b.
Show that y grows above all bounds, which needs y0>0 and f(y0)>0; then at some point u<b, which ensures that u reaches a root in finite time. This root of u is a singularity for y.
The assumptions do not cover at all what the situation is for f(y0)<0.

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