- #1
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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 don't get.
Thanks in advance.
[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 don't get.
Thanks in advance.