Consider the pair of nonlinear differential equations

$F′=FG G′=F2$

As a first approach to a solution for this system, expand the two functions about some nonsingular point in power series:

$F=∑ k=0∞ fk xkk! G=∑ k=0∞ gk xkk!$

Since the right-hand side of both equations contains a product, the denominators in the series will be useful for compactness. The system becomes

$F′ =∑k=1 ∞fk xk-1 (k-1)! =∑ m,l=0 ∞ fmgl xm+l m!l! =FG G′ =∑k=1 ∞gk xk-1 (k-1)! =∑ m,l=0 ∞ fmfl xm+l m!l! =F2$

Equating coefficients of equal powers gives

$fk+1 =∑ m=0k (km) fm gk-m gk+1 =∑ m=0k (km) fm fk-m$

which can be evaluated recursively as desired. The first few coefficients in terms of initial values for each function are

$f1=f0 g0 f2=f0 g02 +f03 f3=f0 g03 +5f03 g0 f4=f0 g04 +18f03 g02 +5f05 f5=f0 g05 +58f03 g03 +61f05 g0 g1 =f02 g2 =2f02 g0 g3 =4f02 g02 +2f04 g4 =8f02 g03 +16f04 g0 g5 =16f02 g04 +88f04 g02 +16f06$

The coefficients in the left-hand column are the beginning of the known series A008971, while those in the right-hand column are the beginning of A008303. Unfortunately this does not particularly help in identifying the underlying analytic solution, as the generating functions given on the linked pages are more complicated than what follows.

When both functions have the same initial condition, ${f}_{0}={g}_{0}$ , the two columns are equal, indicating a common special solution. This solution can be determined with the simple integration

$F′=F2 −1F =x-1f0 F=1 1f0 -x =G$

and can be verified by noting that in this special case ${f}_{n}=n!\phantom{\rule{.2em}{0ex}}{f}_{0}^{n+1}$ , so that the power series expansions are simply geometric series multiplied by ${f}_{0}$ .

The full system has a simple analytic solution due to the existence of an exact constant. Form an equivalence and integrate both sides:

$F′F =F2G =G′G F2 =G2 +c2 → c=f02 -g02$

This exact constant allows the immediately direct integrations

$F′ =FF2 -c2 dF FF2 -c2 =dx 1c sec−1 Fc =x+1c sec−1 f0c → F=csec(cx +sec−1 f0c)$

$G′ =G2+c2 dGG2 +c2 =dx 1c tan−1 Gc =x+1c tan−1 g0c → G=ctan(cx +tan−1 g0c)$

using derivatives of trigonometric functions. It is straightforward to verify that power series expansions of these functions match exactly the forms determined recursively.

It should be noted that the offsets inside the two solution functions represent the same value:

$sec−1 f0c =tan−1 g0c =sin−1 g0f0$

This makes it trivial to substitute the solution back into the original system for confirmation of the result.

It is further clear from the structure of the two direct integrations that both of these functions become the common special solution as $c\to 0$ . This need not be shown explicitly.

For comparison, consider the related pair of nonlinear equations

$F′ =−FG G′ =−F2$

with a simple change of sign on the right-hand sides. The common special solution is trivially different:

$F′ =−F2 1F =x+1f0 F=1 1f0 +x =G$

The exact constant remains the same,

$F′F =−F2G =G′G F2 =G2 +c2 → c=f02 -g02$

while the presence of the minus sign alters the direct integrations:

$F′ =−FF2 -c2 −dF FF2 -c2 =dx 1c csc−1 Fc =x+1c csc−1 f0c → F=ccsc(cx +csc−1 f0c)$

$G′ =−(G2 +c2) −dG G2 +c2 =dx 1c cot−1 Gc =x+1c cot−1 g0c → G=ccot(cx +cot−1 g0c)$

The offsets inside the two solution functions again represent the same value, but different from the previous section:

$csc−1 f0c =cot−1 g0c =cos−1 g0f0$

It is again clear from the structure of the two direct integrations that both of these functions become the common special solution as $c\to 0$ .

It is straightforward to see via expansions that this solution differs from the previous one only in that both functions here are odd in the independent variable x. This can be understood by noting that the right-hand minus signs can be absorbed into this variable, but it is instructive to show it directly.

In order to do so, remember that for two trigonometric functions related by algebraic inversion, their inverse functions are equal with algebraically inverse arguments:

$csc−1x =sin−1 1x sec−1x =cos−1 1x cot−1x =tan−1 1x$

With a negative argument, the sine appearing in the first function of the second solution becomes

$sin( sin−1 cf0 -cx) =cf0 coscx -g0 f0 sincx =cos(cx +cos−1 cf0)$

so that one has

$F(−x) =ccos(cx +cos−1 cf0 ) =csec(cx +sec−1 f0c)$

which is precisely the first function of the first solution. Similarly, with a negative arguments the sine and cosine in the second function of the second solution become

$sin( tan−1 cg0 -cx) =cf0 coscx -g0 f0 sincx =cos(cx +tan−1 g0c) cos( tan−1 cg0 -cx) =g0 f0 coscx +cf0 sincx =sin(cx +tan−1 g0c)$

so that one has

$G(−x) =csin(cx +tan−1 g0c ) cos(cx +tan−1 g0c ) =ctan(cx +tan−1 g0c)$

which is precisely the second function of the first solution.

As a final note, when $|{g}_{0}|>|{f}_{0}|$ all of the exact solutions in this presentation remain purely real. This is most easily understood by recognizing that the change of sign on ${c}^{2}$ in all direct integrations merely replaces inverse trigonometric functions with inverse hyperbolic functions. The initial definition of the exact constant can then be seen as a choice as to which set of functions to use as solutions.