# Are these coefficients swapped in this example, or am I wrong?

I believe I'm staring at an error here. Attached is a screenshot. Am I right in understanding that the author swapped the coefficients around in solving q using the quadratic formula? I can see why the mistake was made, if I'm right about the mistake, because of the name of the coefficients in the differential equation. The issue is that 0.1 was used as a in the quadratic. It should be used as b?

## 1 Answer

The solution of the quadratic equation $$\frac{dP}{dt}=0$$ is correct, taking into account that the choice of coefficients runs counter to the 'usual' parametrization. Note that the denominator in the 'usual parametrization' is $$2a$$, double the coefficient before the quadratic term. This is the same here, where it is called $$2b$$, again double the coefficient before the quadratic term.

If the differential equation is correct with using $$a$$ and $$b$$ as it did I cannot say, as those values miss units, they are just numbers. If you think that the coefficient of $$P^2$$ should be $$-a=-\frac1{10}$$, then the differential equation is incorrect, but the step from differentioal equation to finding the equilibrium points is OK.

• +1 I was going to say that, but I'm late by 5 minutes. Yes, the symbols $a$ and $b$ are swapped compared to usual polynomial of degree $2$, but they are consistently swapped both in the polynomial and in the whole solution, hence the solution is correct. – CiaPan Apr 24 at 21:45
• Thanks, it confused the hell out of me for some reason. But I see now that there is in fact nothing wrong here. Thanks again! – amateurjustin Apr 25 at 7:57