General ODE and rewriting solution

Question

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.

Answer 1

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.

Answer 2

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 $a\neq 0$.