Answer
($\frac{cy}{a}$)+ ($\frac{1}{a}$)(d-($\frac{cb}{a}$)ln(ay+b)=x+K
Work Step by Step
Question: $\frac{dy}{dx}$=$\frac{ay+b}{cy+d}$
1. Separate the equation putting all variables with x on one side and y on the other by cross multiplying:
dy(cy+d)=dx(ay+b)
$\frac{cy+d}{ay+b}$dy=dx
2.Integrate both sides, left side with respect to y and the right side with respect to x. Since $\frac{cy+d}{ay+b}$ cannot simply be integrated, you have to first separate it into partial fractions. So in order to do that, according to the rules of partial fractions since the order of y on the denominator and numerator is the same we have to first proceed with long division as shown in the picture.
According to the rules of long division theorem : 7 divided by 3 will give 2 and a remainder of 1 so:
3 x 2 +1 = 7
so using this we can now get the partial fractions:
$\frac{ay+b}{ay+b}$($\frac{c}{a}$)+$\frac{(d-cb/a)}{ay+b}$=$\frac{cy+d}{ay+b}$
cancelling the two ay+b on the left side we get our partial fractions:
($\frac{c}{a}$)+$\frac{(d-cb/a)}{ay+b}$=$\frac{cy+d}{ay+b}$
3. Now we can easily integrate it and get our final answer(Note that the integration of dx on the right side which is dx is just x because it is the integration of 1 and K is just a constant)
($\frac{cy}{a}$)+($\frac{1}{a}$)(d-$\frac{cb}{a}$)ln(ay+b)=x + K
