- #1

inglezakis

- 7

- 0

ydx+ydy=0, constaint: xy=a

By substituting x with y/a and after some manipulations we arrive to

(-a/y)dy+ydy=0 and on integration we have -aln(y2/y1)+(y2^2-y1^2)/2=0

Suppose we divide the original equation by y. We have

dx+dy=0 and then the solution is (x2-x1) + (y2-y1) =0 (the constraint is eliminated by eliminating y)

Or, we divide by (y^2), we have

dx/y+dy/y=0 and the solution is (x2^2-x1^2)/2a+ln(y2/y1)=0

Here the constraint is needed in order to substitute y for x in the dx term.

So, by dividing the original equation we arrive in different solutions. I don't have the space here, but it can be shown that the 3 solutions above converge for small (x1-x2). Anyway, the issue is when and under what condition we are allowed to divide the differential equation, the purpose being to arrive in the same solution.