Differential Equation and Bernoulli I'm working on the following question with initial condition (x,u)=(2,1):
$$ \frac{du}{dx} = u(2-u)$$
This appears to be a Bernoulli form and substitution v=1/u is needed.
$$ v=1/u $$
$$ v' = \frac{-1}{u^2} \frac{du}{dx} $$
$$ v' = \frac{-1}{u^2} u(2-u) $$
$$ v' = \frac{-1}{u}(2-u) $$
$$ v' = \frac{-2}{u}+1  $$
$$ v' + \frac{2}{u} = 1  $$
$$ v' + 2v = 1  $$
Is this the correct away to approach this problem? It now looks like it is in standard linear form and can be solved by taking the integrating factor. However, I'm not sure what v' represents, $ \frac{dv}{du} $ OR $ \frac{dv}{dx} $?
How do I get the solution in u(x)=... form?
 A: I would rather do it in the naive, but easy way:
$\displaystyle \frac{du}{dx}=u(2-u) \implies \frac{du}{u(2-u)}=dx \implies \int \frac{1}{u(2-u)}du=\int dx$
$$\frac{1}{u(2-u)}=\frac{A}{u}+\frac{B}{2-u}$$
$$A(2-u)+B(u)=1 \implies A=\frac{1}{2}, B=\frac{1}{2}$$
$$\frac{1}{u(2-u)}=\frac{1}{2u}-\frac{1}{2(u-2)}$$
$$x = \frac{1}{2}(\int \frac{1}{u}du - \int\frac{1}{u-2}du )$$
$$ 2x = \ln(\frac{u}{u-2})+C$$
$$ Ce^{2x}=\frac{u}{u-2} \implies Ce^{2x}u-2Ce^{2x}=u \implies u(Ce^{2x}-1)=2Ce^{2x}$$
$$u=\frac{2Ce^{2x}}{Ce^{2x}-1}$$.
Now you can find $C$ by your initial conditions as follows:
$$1=\frac{2Ce^{4}}{Ce^{4}-1} \implies Ce^{4}-1=2Ce^{4} =\implies C=-\frac{1}{e^4}$$
Plugging $\displaystyle C=\frac{-1}{e^4}$ into what we had obtained we'll get:
$$ u=\frac{2e^{2x}}{e^{2x}+e^4} $$
I have to apologize for two things. First of all, I did all these calculations here on MSE because I didn't have a pen and paper around me here, and it was very hard for me to do calculations this way without a pen and paper because I had to do some intermediate steps in my head. So it's possible that I have made some minor mistakes. Secondly, it seems like you were looking for another solution to your problem. Fortunately my method worked, but I can't guarantee this one is the best way to do such problems.
A: You have done the problem correctly. Here, $v' = \frac{dv}{du}$ because you are taking the derivative of both sides of the equation $v=\frac{1}{u}$ with respect to u.
