Separable differential equation

  • Thread starter Syrus
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Homework Statement



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?



Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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7,038

Homework Statement



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?



Homework Equations





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.
 
  • #3
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Thanks Mark. That helped. Can anyone also verify that y=1 is a singular solution to the D.E.?
 
  • #4
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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.
 
  • #5
35,222
7,038
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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