On integration when solving differential equations (specifically separable equations) So here is the differential equation and inititial conditions: $$x \frac{\mathrm{d}y}{\mathrm{d}x}=y(3−y) $$
and $$y(2) = 2$$
 We have to find the equation $y$ in terms of $x ~~[y(x)]$ that is a solution to this differential equation.
Note: Question taken from: Differential equation.
 My question essentially is: can someone explain to me, in a conceptual intuitive sense, why we integrate from $2$ to $x$ in the following derivation? 
So it is can be algebraically manipulated into: $$ \frac{1}{x}~\mathrm{d}x= \frac{1}{y(3−y)}~\mathrm{d}y $$
 Then integrated as: $$\int_2^x \frac{1}{x}~\mathrm{d}x= \int_2^x \frac{1}{y(3−y)}~\mathrm{d}y $$
Which is simplified to $$x^3 = \frac{4y}{3-y}$$The implicit solution.
And then $y$ is isolated and we get:
$$ y(x) = \frac{3x^3}{4+x^3} $$
 Now, as stated at the top, can someone explain to me, in a conceptual intuitive sense, why we integrate from $2$ to $x$?
 A: A lot of short cuts are commonly used in solving differential equations.  These make the working easier, but they do tend to obscure what is really happening.  Maybe it will help if we do things a bit more formally.  Start with
$$\frac{1}{x}= \frac{1}{y(3−y)}\frac{dy}{dx}\tag{$1$}$$
- I haven't split up the $dy$ and $dx$ as it is not clear what, if anything, they mean separately.  (Search math.se pages for lots of discussion and argument on this.)  As the two sides are equal, they remain equal if we perform any legitimate operation on both sides.  Since we want to get rid of the derivative, the operation I choose is integration with respect to $x$, and we get
$$\int_a^b \frac{1}{x}\,dx=\int_a^b \frac{1}{y(3−y)}\frac{dy}{dx}\,dx\ .\tag{$2$}$$
Now how to choose $a$ and $b$?  I will choose $b=x$ since I want an answer in terms of $x$, and I will choose $a=2$ because we are given an initial condition when $x=2$.  So we have
$$\int_2^x \frac{1}{x}\,dx=\int_2^x \frac{1}{y(3−y)}\frac{dy}{dx}\,dx\ .$$
In practice I would actually go from $(1)$ directly here and would not write down $(2)$.  The final step is to recognise the RHS as an example of integration by substitution, so that we have to integrate $1/y(3-y)$ with respect to $y$.  But remember that when we do this we also have to replace the $x$ values in the limits of integration by the corresponding $y$ values.  For the upper limit, if we have an unspecified $x$ value we will replace it by the corresponding (unspecified) $y$ value; for the lower limit, we replace $x=2$ by $y=y(2)=2$.  This gives
$$\int_2^x \frac{1}{x}\,dx=\int_2^y \frac{1}{y(3−y)}\,dy\ .$$
Note that in the first integral equation in your question, the RHS is not correct.  Also, the lower limit of integration $2$ on the RHS is correct but understandably confusing.  It is not $x=2$ as on the LHS but $y=2$, because of the initial condition.  If we had the same differential equation but with initial condition say $y(2)=1$, then the integrals would end up as
$$\int_2^x \frac{1}{x}\,dx=\int_1^y \frac{1}{y(3−y)}\,dy\ .$$
Hope this helps!
A: I have the feeling that, may be, you make life more complex than it is.
Just start with the differential equation $$x \frac{\mathrm{d}y}{\mathrm{d}x}=y(3−y)$$ which is separable and rewrite it as $$\frac{dy}{y(3-y)}=\frac{dx}{x}$$ Integrate both sides (using partial fractions for the lhs). You then find $$\frac{1}{3} \log \Big(\frac{y}{3-y}\Big)=\log(x)+a$$ which rewrite $$ \log \Big(\frac{y}{3-y}\Big)=\log(bx^3)$$ that is to say $$y=\frac{3 b x^3}{b x^3+1}$$ Now, apply the condition $y(2)=2$ which leads to $b=\frac{1}{4}$. Finally, just multiply both numerator and denominator by $4$ to get your good result.
As you see, I only worked with antiderivatives and I used the condition at the end to find the required value for the integration constant.
