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Differential Equation problem

  • #1
I'm asked to find the general solution of the differential equation

[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.
 

Answers and Replies

  • #2
525
6
Well, first of all you can multiply your equation with 2, and rewriting you constant to -(2Ln[C] +1) (which is ofcourse just another constant). After that the other answer more or less rolls out:

[tex]x^2 + 2x + 2ln | x-1 | + y^2 = -(2ln|C| + 1)[/tex]
[tex]x^2 + 2x + 1 + y^2 + 2 ln | x-1 | + 2ln|C| = 0[/tex]
[tex](x+1)^2 + y^2+ 2 ln|C(x-1)| = 0[/tex]
 
  • #3
i see. I never knew you could add numbers to C. Thanx :)
 

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