- #1

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[tex]

x^2dx + y(x-1)dy = 0

[/tex]

I obtained a solution of

[tex]

\frac{1}{2}x^2 + x + ln | x-1 | + \frac{1}{2}y^2 = C

[/tex]

The book, however, gives an answer of

[tex]

(x+1)^2 + y^2 + 2ln |C(x-1)| = 0

[/tex]

I'm sure it's a simplified answer of my own answer. What I don't understand is how a term of +1 and +ln c appeared in the equation after transposing it in the equation. I know for a fact that you can transofrm c into (ln c) since they are constants, but the rest I dont get.

Thanks in advance.