Solving Separable Equations - Get Help Now

[roymkim]

Hi guys,

I was working on this problem regarding separable equations and could not solve it..

dy/dx = ay+b/cy+d

My work:
I reorganized the equation to become dy(cy+d/ay+b) = dx
integrating both sides, you get the integral of (cy+d/ay+b) and dx which is a constant k.
I'm pretty sure that the cy+d/ay+b integration requires long division...but I forgot how to do long division.

I would really appreciate help with this problem~

 

[lurflurf]

I assume you mean
dy/dx = (ay+b)/(cy+d)
You do not need long division just write

ay+b=[a(cy+d)+(bc-ad)]/c for c!=0

(ay+b)/(cy+d)=(ay+b)/d for c=0

 

[roymkim]

The answer to the problem comes out to be

x= (c/a)y + [(ad-bc)/(a^2)]*ln|ay+b|+k; a≠0, ay+b≠0

Can you explain more about how you reorganized the equation to become
ay+b=[a(cy+d)+(bc-ad)]/c for c!=0

(ay+b)/(cy+d)=(ay+b)/d for c=0

what do you mean by c! and what do you mean by c?

 

[HallsofIvy]

c!= 0 is "computer talk" for "c not equal to 0".
As to what lurflurf "means by c", I presume the same thing you did- the coefficient of y in the denominator!

[tex]\frac{dy}{dx}= \frac{ay+ b}{cy+ d}[/tex]
[tex]\frac{cy+ d}{ay+ b}dy= dx[/tex]

Personally, I think I would use "long division" here:
[tex]\frac{cy+ d}{ay+ b}= \frac{c}{a}+ \frac{ad- bc}{a}\frac{1}{ay+ b}[/tex]

And that should be easy to integrate.
(let u= ay+ b)

 

[lurflurf]

yes c is the c you used in dy/dx = (ay+b)/(cy+d). c!=0 means c is not zero, that allows us to divide by it. All the ways to divide do the same things, maybe review them to see this. There are synthetic division, long division, division by inspection, and rearranging the equation like I did.