Equation

DavidD

20th April 2008 - 10:09 AM

How to solve such equation?:
y=(x^2)/2+y*ln|x|

Particularly I stuck on
soving this diferential equation
y'x-x^2-y=0.
I solve it like this:
dy/dx=x-y/x,
S dy=S(x-y/x)dx,
y=(x^2)/2-y*ln|x|+C.

And my solution is right, but in textbook solution is y=x^2+C*x - the more "precisly" solution. So I wonder how was goten such solution and also want to know how to solve equation above.

DavidD

21st April 2008 - 12:32 PM

Ou well, I solve and get x^2+Cx,
Wach out
y'x-x^2-y=0,
y=ux, y'=u'x+u,
(u'x+u)x-x^2-ux=0,
u'x+u-x-u=0,
u'x=x,
(du/dx)*x=x,
du/dx=1,
du=dx,
Sdu=Sdx,
u=x+C,
y/x=x+C,
y=x^2+Cx.

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