I have the system of ode $\frac{da}{dt}=2ab$, $\frac{db}{dt}=b^2-a^2+1-\frac{a}{\pi}$
When I am plotting it in xppaut(see the picture attached). I can see that a quantity is conserved but I am unable to figure out the expression. Can anyone please help? If you can give me the steps on how you get the conserved quantity that would be very helpful as well.
My try: I was using trial and error with $\alpha a^2+\beta b^2$ but didn't get much help. Now I am trying to find out the integrating factor of $\frac{da}{db}=\frac{2ab}{b^2-a^2+1-\frac{a}{\pi}} \Rightarrow (b^2-a^2+1-\frac{a}{\pi})da-(2ab)db=0$. Now this is not exact[$2b\neq -2b$].
Then I got the integrating factor $=\frac{1}{a^2}$ and got the solution $$\frac{b^2}{a}+\frac{1}{a}+a+\frac{\ln a}{\pi}=C$$ for some integrating constant $C$.
Now I am trying to use it somehow. If you get a conserved quantity please let me know.
$$\left(b^2-a^2+1-\frac{a}{\pi}\right)da-(2ab)db=0$$ $$\implies b^2da-(2ab)db=\left(a^2+\frac{a}{\pi}-1\right)da$$ $$\implies \frac{1}{\left(\frac{a}{b^2}\right)^2}d\left(\frac{a}{b^2}\right)=\frac{1}{a^2}\left(a^2+\frac{a}{\pi}-1\right)da$$ $$\implies \frac{b^2}{a}+\frac{1}{a}+a+\frac{\ln a}{\pi}=C$$ Since $C$ is independent of the variable $t$, it is a conserved quantity.