Solving $\frac{dy}{dx} = \frac{ay+b}{cy+d}$ I'm on the section of my book about separable equations, and it asks me to solve this:
$$\frac{dy}{dx} = \frac{ay+b}{cy+d}$$
So I must separate it into something like: $f(y)\frac{dy}{dx} + g(x) = constant$
*note that there are no $g(x)$
but I don't think it's possible. Is there something I'm missing?
 A: You can always rearrange it as follows:
$$\frac{dy}{dx} = \frac{ay+b}{cy+d}$$
$$\frac{cy+d}{ay+b}\frac{dy}{dx} = 1$$
It is clearly separated and you can integrate with respect to $x$ to get:
$$\int\frac{cy+d}{ay+b}\frac{dy}{dx}dx = \int dx$$
$$\int\frac{cy+d}{ay+b}dy = x$$
The left hand side requires partial fractions to complete.
Edit Added due to comment about partial fractions.
Firstly carry out the division so the numerator has smaller degree than the denominator.
$$\int\frac{cy+d}{ay+b}dy = x$$
$$\int\frac{\frac{c}{a}ay+d}{ay+b}dy = x$$
$$\int\frac{\frac{c}{a}ay+\frac{c}{a}b+d-\frac{c}{a}b}{ay+b}dy = x$$
$$\int\frac{\frac{c}{a}(ay+b)+d-\frac{c}{a}b}{ay+b}dy = x$$
$$\int\frac{\frac{c}{a}(ay+b)}{ay+b}+\frac{d-\frac{bc}{a}}{ay+b}dy = x$$
$$\int\frac{c}{a}+\frac{d-\frac{bc}{a}}{ay+b}dy = x$$
$$\frac{c}{a}y+(d-\frac{bc}{a})\frac{\log(ay+b)}{a}=x+K$$
$$\frac{c}{a}y+\frac{(ad-bc)\log(ay+b)}{a^2}=x+K$$
Aside: with separable DEs it is probably better to think about how to rearrange it into the form:
$$f(y)\frac{dy}{dx}=g(x)$$
This is similar to your form but there is no need to have a distinct constant as it can always be considered part of $g(x)$.
A: Consider: $$\frac{dy}{dx} = \frac{ay+b}{cy+d}$$
which can be seen as the following:
\begin{align}
1 &= \frac{c y + d}{a y + b} \, \frac{dy}{dx} \\
&= \frac{c}{a} \, \left[ 1 + \left(\frac{d}{c} - \frac{b}{a} \right) \, \frac{a}{ay + b} \right] \, \frac{dy}{dx}
\end{align}
which becomes
$$\frac{a}{c} \, dx = \left[ 1 + \left(\frac{d}{c} - \frac{b}{a} \right) \, \frac{a}{ay + b} \right] \, dy $$
and leads to, after integration,
$$y + \left(\frac{d}{c} - \frac{b}{a}\right) \, \ln(a y + b) = \frac{a \, x}{c} + \mu_{0},$$
where $\mu_{0}$ is the constant of integration.
A: For your specific example, extracting the horizontal asymptote you can also write $$\frac{cy+d}{ay+b}=\frac c a+\frac{a d-b c}{a (a y+b)}$$ Then $$I=\int\frac{cy+d}{ay+b}\,dy=\frac c a \int dy+\frac{a d-b c} a\int \frac {dy} { (a y+b)}=\frac c a \int dy+\frac{a d-b c} {a^2}\int \frac {a\,dy} { (a y+b)}$$ I am sure that you can take it from here.
