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spazzter08dyk2n 2022-05-17 Answered
We have y = x / y, which is a first-order homogeneous differential equation.
It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that y = ± x 2 + c .
Now if we use the substitution y = u x y = u x + u ,, and rewrite the differential equation as
u x + u = 1 u
and then rearrange to
( 1 1 / u u ) d u = ( 1 x ) d x
by integrating both parts we get that
(1) 1 2 ln | u 2 1 | = ln | x | + c
For y = ± x (a special solution for c=0) u = ± 1, and by plugging ± 1 into (1) we get that
(2) ln | 0 | = ln | x | + c
What does equation (2) mean? ln | 0 | is undefined. Is this of any significance?
Edit 1:
As pointed out when rearranging from u x + u = 1 u to ( 1 1 / u u ) d u = ( 1 x ) d x, we implicitly assumed that u ± 1. Equation (1) does not hold for u = ± 1
Edit 2:
Solving equation (1) for u with u ± 1, we arrive at the same family of equations but with c 0. The fact that c can be zero comes from setting u = ± 1
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Answers (1)

Cristal Obrien
Answered 2022-05-18 Author has 16 answers
When you rearranged your equation to get
( 1 1 / u u ) d u = ( 1 / x ) d x
you have implicitly assumed that u ± 1 so u = ± 1 is out of question.
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