- #1
inglezakis
- 7
- 0
The point of my question is that when we divide a differential equation by a function or variable we result in different solution (not always). Take the example:
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.
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.