General ODE and rewriting solution When considering the general form (which is an initial value problem)
$$\frac{{dy}}{{dt}} = ay - b
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca
% WGKbGaamyEaaqaaiaadsgacaWG0baaaiabg2da9iaadggacaWG5bGa
% eyOeI0IaamOyaaaa!3E8D!
$$
with initial condition y(0)=y0 (Where y0 is an arbitrary initial value)
If
$$\begin{array}{l}a \ne 0\\y \ne \frac{b}{a}\end{array}
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGHb
% GaeyiyIKRaaGimaaqaaiaadMhacqGHGjsUdaWcaaqaaiaadkgaaeaa
% caWGHbaaaaaaaa!3E06!
$$
The testbook I have rewrites the general form as:
$$\frac{{\frac{{dy}}{{dt}}}}{{y - (\frac{b}{a})}} = a
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaada
% WcaaqaaiaadsgacaWG5baabaGaamizaiaadshaaaaabaGaamyEaiab
% gkHiTiaacIcadaWcaaqaaiaadkgaaeaacaWGHbaaaiaacMcaaaGaey
% ypa0Jaamyyaaaa!40EC!
$$
I don't understand why they would rewrite in this way. The only connection I can make in my mind that the derivative is related to the limit which 1/0 would be undefined or a condition associated with a limit. Any insight that some one can provide for this rewrite would really clear up a lot for me. 
This leads to 
a solution of the initial value problem of 
$$y = (\frac{b}{a}) + [y0 - (\frac{b}{a})]{e^{at}}
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr
% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9
% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x
% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2
% da9iaacIcadaWcaaqaaiaadkgaaeaacaWGHbaaaiaacMcacqGHRaWk
% caGGBbGaamyEaiaaicdacqGHsislcaGGOaWaaSaaaeaacaWGIbaaba
% GaamyyaaaacaGGPaGaaiyxaiaadwgadaahaaWcbeqaaiaadggacaWG
% 0baaaaaa!46A3!
$$
Thanks in advance.
 A: The reason that the ODE is written as you showed is so that terms containing $y$ are on one side of the equation only and terms not containing $y$ are also on one side of the equation only. Rewriting that way allows you to integrate both sides of the equation, the left hand side with respect to $y$ and the right hand side with respect to $t$. The left hand side becomes $\ln (y - \frac{b}{a})$, while the right hand side becomes $at + C$, where $C$ is some constant.
A: Essentially you want to integrate the ODE  $y’(t)=ay(t)-b$ by t in order to obtain the solution $y(t)$. The problem with doing this directly is that you do not know the integral of the right side since $y(t)$ is unknown yet. Therefore we bring all $y(t)$ on one side through dividing by the right side, leading to
$$\frac{y’(t)}{ay(t)-b}=1.$$
This step is called separation of variables. This is one approach of solving ODEs. Of course we assumed that $ay(t)-b\neq 0$. Now we can integrate both sides with regards to $t$. On the left side we make the substitution $z:=ay(t)-b$. Assuming $a\neq 0$ integration by parts then yields
$$\frac1 a\ln\lvert(ay(t)-b)/(ay_0-b)\rvert=t.$$
Finally this gives the solution
$$y(t)=b/a + (y_0-b/a)e^{at}.$$
By plugging it back into the ODE we see that the only assumption we need is $a\neq 0$.
