Here's the code to study the equation in Mathematica:

```
[tex]ode=z''[b](6(1 - f)z[b] + (1 + f) b z'[b]) ==\\
(15-9f)(z'[b])^2+\frac{2(1-f)z[b]z'[b]}{b}+\frac{4f(z'[b])^{5/2}}{b^{1/2}}[/tex]
bstart = choose something;
bend = 1.2;
zderivstart = 0.01;
zstart = 0.01;
plist = Table[{0}, {10}];
f = 0;
For[i = 1, i <= 10, i++,
f += 0.1;
sol1 =
NDSolve[{ode, z[bstart] == zstart, z'[bstart] == zderivstart},
z, {b, bstart, bend}];
fn1[t_] := Evaluate[z[t] /. Flatten[sol1]];
plist[[i]] =
Plot[fn1[t], {t, bstart, bend}, PlotRange -> {{0, bend}, {0, 2}}
];
];
fin1 = Show[{plist}, PlotLabel -> {StyleForm["b0=chose here", FontSize -> 16]}]
```

Case 1: z(0.3)=0.01, z'(0.3)=0.01

As f is varied from 0.1 to 1, the derivative grows larger and larger until a vertical asymptote is reached. For this case, this occurrs at about b=2. This is exhibited in the first plot.

Case 2: z(0.2)=0.01, z'(0.2)=0.01

The same behavior is encountered however, the derivative is increasing faster. In this case the vertical asymptote is not reached until f=1 and b is about 1.1 (neglect the vertical line near 1).

Case 3: z(0.1)=0.01, z'(0.1)=0.01

As f is varied from 0.1 to 0.9, the plot rises faster and faster until at f=0.5, a vertical asymptote is encountered at around 1 and is a function of f. That is, when f=0.5, the asymptote is reached at about 1.14, when f=0.6, it's about 1.0, when f=0.7, its 0.95, and so on for increasing f. The first group of plots exhibits this behavior (the vertical lines are spurious and are not to be considered).

The trend seems to become more severe as b0 approaches 0. My suspicion is that perhaps a delta function satisfies the equation at b0=0. Often in cases like these it's crucial to have an understanding of what the equation models in real life. Does the behavior exhibited below conform to a real-world application and do delta functions have any relevance?

Edit: Perhaps it's not correct to say "a vertical asymptote is reached". Mathematica just encounters a slope with a very high value.