Question

Consider the differential equation dy/dx = (y - y^2)/x.

Verify that y = x/(x + C) is a general solution for the given differential equation and show that all solutions contain (0,0).

Answer 1

This is a separable DE.

Answer 2

I tried integrating dy/dx and differentiating y, but I can't seem to be able to show that...

Answer 3

Rewrite it as:
\[\frac{dy}{y(1-y)}=\frac{dx}{x}\]
Now, integrate both sides.

Answer 4

Can you do the integration?

Answer 5

Yes, but I get something with arctangent and a natural log.

Answer 6

You can verify it is a solution just by plugging in the form for y into the DE and differentiating appropriately.

Answer 7

You better avoid the arctan. Here:
\[\int\limits \frac{dy}{y(1-y)}=\int\limits ({2 \over y}-{2 \over y-1})dy=2\ln|{y \over y-1}|+c\]
So, the solution of the DE is:
\[2\ln|{ y \over y-1}|=\ln|x|+c.\]
Now, simplify!

Answer 8

Oh crap! Do as TT said, just plug the given solution and verify that it satisfies the given DE.

Answer 9

y = x/(x + C)
into
dy/dx = (y - y^2)/x.
gives
\[\frac{x+C-x}{(x+C)}^2=\frac{\frac{x}{x+C}-(\frac{x}{(x+C)})^2}{x}\]it is not very hard to show that both sides are the same algebraically. Why do we need to integrate?

Answer 10

Yeah, TT's way is much simpler. No need to integrate :P
Thanks both of you :)

Answer 11

No problem! :D

Answer 12

as far as showing that (0,0) is int there that seems to come from simplifying the above, but I'm not sure...

Answer 13

*is in there

Answer 14

\[y(x)=\frac{x}{e^c+x} \]
http://www.wolframalpha.com/input/?i=+y%27%3D%28y+-+y%5E2%29%2Fx

Answer 15

typo
plugging in gives\[\frac{x+C-x}{(x+C)^2}=\frac{\frac{x}{x+C}-(\frac{x}{(x+C)})^2}{x}\]