Questions on the differential equation $df/dx=-[f(x)]^2$ I have another group project problem
 I am having trouble with.
Here is the first part of the problem:
"Consider the differential equation $df/dx=-[f(x)]^2$, with initial condition $f(0)=a$."
"Express $f(x)$ as the power series $\sum_{n=0}^\infty c_nx^n$"
"a) What is $c_0$?"
"b) What is the power series $f'(x)$?"
I believe I solved the equation correctly but I am not sure where to go from there.
Solving for $f(x)$ I get: $f(x)=\frac{1}{x-c}$
Then solving for the initial condition I arrive at: $a=1/c$, which gives me $f(x)=\frac{1}{x-1/a}$.
I am just not sure how to express $f'(x)$ or $f(x)$ as a power series...
Any help would be greatly appreciated.
 A: I think that one want that you use power series to find $f$. 
a) if $\displaystyle f(x)=\sum_{n\geq 0} c_n x^n$, then you have $c_0=f(0)=a$.
b) You have $\displaystyle f^{\prime}(x)=\sum_{n\geq 1} nc_n x^{n-1}$.
Now $\displaystyle f(x)^2=\sum_{n\geq 0} (\sum_{k=0}^n c_kc_{n-k})x^n$.
So you get that for $n\geq 0$
$$(n+1)c_{n+1}=-\sum_{k=0}^n c_kc_{n-k}$$
Using this formula gives you $c_1=-a^2$, $c_2=a^3$. Now prove by induction that $c_n=(-1)^n a^{n+1}$ for all $n$. 
A: I think $f(x)=1/(x+1/a)=a/(1+ax)$
Recall the formula for the geometric series $$\sum_{k=0}^{\infty}y^k=\frac1{1-y}$$
A: Nota Bene:  Throughout this answer, I have deviated somewhat from the exact statement of the question as asked by allowing the initial condition $f(x_0) = a$ for arbitrary $x_0 \in \Bbb R$; $x_0 = 0$ is of course then seen as a very special case.
First of all, the relation 'twixt $a$ and $c$ as given in the text of the question has a small but significant error.  Before going into the details, let's check the general form of the proposed solution
$f(x) = (x - c)^{-1}; \tag{1}$
we then have
$f'(x) = -(x - c)^{-2} = -[f(x)]^2, \tag{2}$
which shows that (1) is the correct form of the solution.   (1) may be derived from
$f'(x) = -[f(x)]^2 \tag{3}$
as follows:  assuming for the moment that $f(x) \ne 0$, (3) may be written as
$-[f(x)]^{-2} f'(x ) = 1; \tag{4}$
that is
$[[f(x)]^{-1}]´ = -[f(x)]^{-2}f'(x) = 1, \tag{5}$
and upon integrating (5) with respect to $x$ we find
$\int_{x_0}^x [[f(s)]^{-1}]' ds = \int_{x_0}^x ds, \tag{6}$
or
$[f(x)]^{-1} - [f(x_0)]^{-1} = (x - x_0), \tag{7}$
and if (7) is re-arranged we obtain
$f(x) = ((x - x_0) + [f(x_0)]^{-1})^{-1}. \tag{8}$
If we are given that $f(x_0) = a$, then (8) may be written
$f(x) = ((x - x_0) + a^{-1})^{-1} = \dfrac{a}{a(x - x_0) + 1}. \tag{9}$
In deriving (9), we have made use of the assumption that $f(x) \ne 0$; this hypothesis is warranted by the fact that, since $-[f(x)]^2$ is Lipschitz continuous (locally) in $f(x)$, the solutions to (3) are (locally) unique; thus the only solution such that $f(x_1) = 0$ for some $x_1$ is the solution $f(x) = 0$; this in turn implies that any solution taking a value $f(x_1) \ne 0$ at some $x_1$ can never take the value $0$ as long as it is defined; so we are safe in the assumption that $f(x) \ne 0$ in the above derivation of (9).
If we compare (9) to the OP's solution
$f(x) = \dfrac{1}{x - a^{-1}}, \tag{10}$
we see that (10) yields
$f(0) = -a, \tag{11}$
in contrast to
$f(0) = a \tag{12}$
which follows from (9), setting $x_0 = 0$.  The discrepancy 'twixt (11) and (12) indicates our OP's error previously mentioned; it doesn't seem to have infected the answers of either Michael or Kelenner, and whether or not our OP FofX's attempts were hindered by this error I cannot say.
Proceding, we note that if $\vert a(x - x_0) \vert < 1$, we have the convergent series
$\dfrac{1}{1 + a(x - x_0)} = \sum_0^\infty (-a(x - x_0))^k, \tag{13}$
whence
$f(x) = \dfrac{a}{1 + a(x - x_0)} = a\sum_0^\infty (-a(x - x_0))^k$
$= \sum_0^\infty (-1)^k a^{k + 1}(x - x_0)^k = \sum_0^\infty c_k (x - x_0)^k. \tag{14}$
We can read the coefficients $c_k$ off from (14); we see that
$c_0= a, \tag{15}$
in accord with the answer given by Kelenner.  We also easily see that
$c_k = a(-a)^k = (-1)^k a^{k + 1}, \tag{16}$
which K. also obtained.  The power series for $f'(x)$ may be had via term-by-term differentiation of (14); this is allowed since (14), being a geometric series with ratio $-a(x - x_0)$, converges absolutely and uniformly for $\vert a(x - x_0) \vert  < 1$, that is, for $\vert x - x_0 \vert < \vert a^{-1} \vert$.  The result is then
$f'(x) = \sum_0^\infty (-1)^k k a^{k + 1} (x - x_0)^{k - 1}. \tag{17}$
Finally, we should observe that the radius of convergence of the infinite series (13) at $x_0$ is in fact $\vert a \vert^{-1}$.  If $\vert x - x_0 \vert \ge \vert a \vert^{-1}$, then the ratio test will fail when applied to (13), and in any event,  $\vert a(x - x_0) \vert > 1$ and the terms of the series increase without bound as $k \to \infty$.  Though the radius of convergence of these series at $x_0$ is in fact $\vert a \vert^{-1}$, it is still the case that the function $f(x)$ defined by (9) is well-behaved for arbitrarily large $x$ provided $a > 0$, and for arbitrarily large negative $x$ when $x < 0$; but when $a > 0$ a singularity in the solution $f(x)$ is encountered as $x \to x_0 - a^{-1}$ from above; $f(x) \to \infty$ under such circumstances.  Likewise for $a < 0$, $f(x) \to -\infty$ as $x \to x_0 - a^{-1}$ from below.  Thus we see that, for $\vert a \vert \ne 0$, the solution may always be continued indefinitely on one side of $x_0$ or the other, depending upon the sign of $a$, but the series expansion only holds for $\vert x - x_0 \vert < \vert a \vert^{-1}$.
And for a final "finally", let us take a final look at the special case $a = 0$, pointed out by Christian Blatter in his comment.  Certainly formulas such as (9) or (10) are invalid when $a = 0$, as is (7), result of the integral in (6).  But these difficulties are circumvented via the realization that $f(x) = 0$ is the unique solution with $f(x_0) = a = 0$, as has been discussed in the above.  In this case of course the series has infinite radius of convergence, corresponding (intuitively, at least) to the fact that $\vert a \vert^{-1} \to \infty$ as $a \to 0$.  Indeed, we have $c_k = 0$ for every value of the index $k$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
