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Separable differential equation

by Syrus
Tags: differential, equation, separable
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Syrus
#1
Feb4-13, 02:29 PM
P: 202
1. The problem statement, all variables and given/known data

I am asked to find a singular solution of the D.E. dy/dx = (xy+2y-x-2)/(xy-3y+x-3). I am first solving to find the general solution form of the D.E., and so far have it to:

[(x+2)/(x-3)]dx = [(y+1)/(y-1)]dy

From here, of course, you integrate both sides, but I am struggilng to find the best technique of integration. Any ideas?



2. Relevant equations



3. The attempt at a solution
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Mark44
#2
Feb4-13, 02:48 PM
Mentor
P: 21,216
Quote Quote by Syrus View Post
1. The problem statement, all variables and given/known data

I am asked to find a singular solution of the D.E. dy/dx = (xy+2y-x-2)/(xy-3y+x-3). I am first solving to find the general solution form of the D.E., and so far have it to:

[(x+2)/(x-3)]dx = [(y+1)/(y-1)]dy

From here, of course, you integrate both sides, but I am struggilng to find the best technique of integration. Any ideas?



2. Relevant equations



3. The attempt at a solution
You can use polynomial long division on each of your two rational expressions, or you can add and subtract the same quantity in each numerator so as to get expressions that are easier to work with. Both techniques produce the same results.

For example, (x + 2)/(x - 3) = (x - 3 + 5)/(x - 3) = (x - 3)/(x - 3) + 5/(x - 3) = 1 + 5/(x - 3). The same sort of idea works with the other rational expression.
Syrus
#3
Feb4-13, 05:42 PM
P: 202
Thanks Mark. That helped. Can anyone also verify that y=1 is a singular solution to the D.E.?

Syrus
#4
Feb4-13, 10:43 PM
P: 202
Separable differential equation

I have solution, which I disagree with, claiming y=0 is a singular solution since, upon substitution, it doesn't seem to produce an identity.
Mark44
#5
Feb5-13, 12:22 AM
Mentor
P: 21,216
Quote Quote by Syrus View Post
I have solution, which I disagree with, claiming y=0 is a singular solution since, upon substitution, it doesn't seem to produce an identity.
Which should suggest that y = 0 isn't a solution at all.

If y ##\equiv## 0 is a (purported) solution, then it follows that dy/dx ##\equiv## 0. However, if y = 0, from the diff. equation, we have dy/dx = (-x - 2)/(x - 3), which is zero only if x = -2.


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