Bernoulli equation when $y$ can be 0 (e.g. Tom Apostol's calculus, vo1, ex $8.5:16$) In the Section $8.5$ of his book, in exercise $13$, Tom Apostol gives a way to solve Bernoulli equation of form:
$y' + P(x)y = Q(x)y^n$
When $y \ne 0$, by solving the linear equation $v' + kP(x)v = kQ(x)$, where $k = 1-n$ and $y = f^k(x) = g(x) = v$.
That is fine, and relatively easy to prove, but just above the exercise $13$ (without proving it) he wrote that we can always transform the Bernoulli equation to a linear first-order equation, and the proofs I found online implicitly assumed that $y \ne 0$.
Furthermore in his excercise $16$, he explicitly wrote that the initial condition is that $f(1) = 0$, and a proposed solution just applied the result as if $f(x) \ne 0$ on the whole interval:
$\tag{16} xy' - 2y = 4x^3 y ^\frac{1}{2}$
Why is it justified to ignore that $f(1) = 0$? Can someone point me to the proof of the equation equivalence mentioned above, which is valid in case when $y = 0$?
Note: This could be seen as a follow up to my previous question, which also discusses a case when the differential equation is not-defined at a point, but the solution is.
 A: $$
a(x)y' + b(x)y + c(x)y^n = 0
$$
Assuming that
$$
\frac{b(x)}{a(x)}, \frac{c(x)}{a(x)}
$$
always exist for all $x$
We can write as
$$
y' + P(x)y + Q(x)y^n = 0
$$
we have
$$
y = v^k
$$
then
$$
kv^{k-1}v' + P(x)v^k + Q(X)v^{kn} = 0
$$
or
$$
kv' + P(x)v + Q(x)v^{kn - k + 1} =0
$$
you can always choose a $k$ such that the following holds
$$
kn - k + 1 = k(n-1) + 1 = 0 \implies k = \frac{1}{1 -n}
$$
you may ask what about $n=1$ well this is linear by definition.
$$
y' + P(x)y + Q(x)y^1 = y' + (P(x)+ Q(x))y = 0
$$
The factorising $v^{k-1}$ is explicitly given by
$$
v^{k-1}\left[kv' + P(x)v + Q(X)v^{kn - k + 1} \right] = 0
$$
the trivial solution is $v = 0 \implies y = 0$ which you get for free from the initial equation.
You may ask what about $k < 1$ then we then have
$$
\frac{1}{1 -n} < 1 \implies n < 0
$$
so if $k < 1$ thus $v^{k-1}$ is not defined for $v=0$ we also have $n < 0$ which means $y^n$ is not defined.
If $n < 0$ we can use $v^{-k}$
$$
y' + P(x)y + Q(x)y^{-n} \implies -(k + 1)v^{-k-1}v' + P(x)v^{-k} + Q(x)v^{kn}
$$
so we have
$$
v^{-k-1}\left[-(k + 1)v' + P(x)v + Q(x)v^{kn + k + 1}\right]
$$
we then have
$$
kn + k + 1 = 0 \implies k = -\frac{1}{1 + n}
$$
