Help needed in solving a system of DE The system of DE is:
$$\frac{dI}{db}=-\frac{b}{c}\frac{dJ}{db}-\frac{2ab+1}{2c}J$$
$$\frac{dJ}{db}=\frac{b}{c}\frac{dI}{db}-\frac{2ab-1}{2c}I$$
Assume that $a$ and $c$ are constants and both $I$ and $J$ tends to zero as $b$ tends to $\infty$.

To be honest, I have never solved a system of DE before. I usually use W|A if I am in a situation like this but in this case, even W|A fails. I am not sure if it is even possible to solve the above system. 
Any help is appreciated. Many thanks!

Removing mixed derivative terms is possible by inserting the second equation's $dJ/db$ into the first equation and vice versa:
$$
\frac{dI}{db} = -\frac{b^2}{c^2} \frac{dI}{db} + 
\frac{b}{c}\frac{2ab-1}{2c} I -
\frac{2ab+1}{2c} J \\
\frac{dJ}{db} = -\frac{b^2}{c^2} \frac{dJ}{db} -
\frac{2ab-1}{2c} I -
\frac{b}{c}\frac{2ab+1}{2c} J
$$
rearranging gives two coupled equations
$$
\frac{dI}{db} + 
\left(-\frac{c^2}{c^2+b^2}\frac{b}{c}\frac{2ab-1}{2c}\right) I = 
\left(-\frac{c^2}{c^2+b^2}\frac{2ab+1}{2c}\right) J \quad (*) \\
\frac{dJ}{db} +
\left(\frac{c^2}{c^2+b^2}\frac{b}{c}\frac{2ab+1}{2c}\right) J =
\left(-\frac{c^2}{c^2+b^2}\frac{2ab-1}{2c}\right) I \quad (**)
$$
which one can turn into a vector equation:
$$
\frac{d}{db}
\left[
\begin{array}{c}
I \\
J
\end{array}
\right]
=
\left[
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right]
\left[
\begin{array}{c}
I \\
J
\end{array}
\right]
$$
with the matrix
$$
A =
\left[
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right]
=
\left[
\begin{array}{cc}
\frac{b}{c^2+b^2} \frac{2ab-1}{2} & 
\frac{c}{c^2+b^2} \frac{2ab+1}{2} \\
-\frac{c}{c^2+b^2} \frac{2ab-1}{2} &
-\frac{b}{c^2+b^2} \frac{2ab+1}{2}
\end{array}
\right]
$$
This is now cast into a form like it is used for dynamical systems, there could be useful tools to get more analytic information.
A graph of the directional vector field would be nice now, with stationary points etc.
The case $a = c = 1$
Here the matrix reduces to
$$
A =
\left[
\begin{array}{cc}
\frac{b}{1+b^2} \frac{2b-1}{2} & 
\frac{1}{1+b^2} \frac{2b+1}{2} \\
-\frac{1}{1+b^2} \frac{2b-1}{2} &
-\frac{b}{1+b^2} \frac{2b+1}{2}
\end{array}
\right]
$$
This is the given candidate for $I$:
$$
I^c(b) = 
\frac{e^{-\sqrt{1 + b^2}}}{\sqrt{1 + b^2}\sqrt{b + \sqrt{b^2+1}}}
$$
with derivative
$$
\frac{dI^c}{db}(b) =
-\frac{e^{-\sqrt{1+b^2}} 
\left(2 b^3 + \left(2 \sqrt{1+b^2} + 3 \right) b^2 +
\left(3 \sqrt{1+b^2}+2\right) b + 1 \right)}
{2 (1 + b^2)^{3/2} \left(\sqrt{1+b^2}+ b \right)^{3/2}}
$$
from equation $(*)$ we get
$$
J^c(b) = \frac{2(1+b^2)}{2b+1}\left(
-\frac{e^{-\sqrt{1+b^2}} 
\left(2 b^3 + \left(2 \sqrt{1+b^2} + 3 \right) b^2 +
\left(3 \sqrt{1+b^2}+2\right) b + 1 \right)}
{2 (1 + b^2)^{3/2} \left(\sqrt{1+b^2}+ b \right)^{3/2}}
- \frac{2(1+b^2)}{b(2b-1)}
\frac{e^{-\sqrt{1 + b^2}}}{\sqrt{1 + b^2}\sqrt{b + \sqrt{b^2+1}}}
\right)
$$
