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 xby 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 = ± 1Edit 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

Question

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  xby 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  =  ±  1Edit 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

Cristal Obrien · Accepted Answer

When you rearranged your equation to get  (      1          1              /            u      −      u        )  d  u  =  (  1      /    x  )  d  xyou have implicitly assumed that   u  ≠  ±  1 so   u  =  ±  1 is out of question.

We have y ′ </msup> = x <mrow class="MJX-TeXAtom-ORD"> / </mrow>

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