Question

This is the final part of a DE problem, i worked out the whole problem but i just do not know how i can solve for y in this case:
(y/x)^2-(y/x)=ln(x)+C

Answer 1

hmm, let v = y/x, or vx = y

Answer 2

ok, i still dont see how i could solve it:\[v ^{2}-v=\ln(x)+C\]

Answer 3

it might help to know how this all started

Answer 4

Here is the original problem:
Solve the DE:\[(x^2- xy + 2y^2 )dx + (x^2 - 2xy)dy = 0\]

Answer 5

is it an exact DE?

Answer 6

no, it fails the test

Answer 7

bummer ... :) so your intent was to make it separable right?

Answer 8

yeah, pretty much then just solve for y

Answer 9

\[(x^2- xy + 2y^2 )dx + (x^2 - 2xy)dy = 0\]
\[(x^2 - 2xy)dy = (x^2- xy + 2y^2 )dx\]
\[(x - 2y)dy = (x- y + \frac{2}{x}y^2 )dx\]
\[\frac{dy}{dx} = \frac{(x- y + \frac{2}{x}y^2 )}{(x - 2y)}\]
\[y' = \frac{1- \frac yx + 2~(\frac yx)^2 )}{1 - 2\frac yx}\]
\[y=vx~:~y'=v'x+v\]
\[\frac{dv}{dx}x+v = \frac{1- v + 2~v^2}{1 - 2v}\]
solve for dv/dx

Answer 10

can we factor the right side any?

Answer 11

\[\frac{dv}{dx}x+v = \frac{1- v + 2~v^2}{1 - 2v}\]
\[\frac{dv}{dx}x = \frac{1- v + 2~v^2}{1 - 2v}-v\]
\[\frac{dv}{dx}x = \frac{1- v + 2~v^2-v+2v^2}{1 - 2v}\]
\[\frac{dv}{dx}x = \frac{1- 2v + 4~v^2}{1 - 2v}\]
\[\frac{1 - 2v}{1- 2v + 4~v^2}{dv} = \frac 1xdx\]
the left might be a good candidate for partial decomp ...

Answer 12

Isn't it possible to treat it like a quadratic?

\[\frac{ y^{2} }{ x^{2} }-\frac{ y }{ x }=lnx + C\]
\[y^{2}- xy - x^{2}(lnx + C) = 0\]
Quadratic formula:
\[\frac{ x \pm \sqrt{x^{2} - 4(-x^{2}lnx+Cx^{2}}) }{ 2 }\]

? Just an idea.

Answer 13

oh, that makes sense, i think thats the correct approach

Answer 14

I just noticed it was quadratic in y, so seemed logical.

Answer 15

im still trying to determine how it got to that form :)

Answer 16

Haha. Everything bu y is treated as a constant. So a is 1, b is -x, c is -x^2(lnx+C)

Answer 17

but yes, the soluton from wolf is very quadratic formulaish looking

Answer 18

Might be able to just simplify it from where we have it then and accept that. Of course an initial condition would be nice so we don't have a C in there.

Answer 19

yeah, its usually nice to get rid of the constant, but no initial condition was given in this problem.

Answer 20

Oh well, so we just deal with the C in there then and simplify it out : )

Answer 21

\[y = \frac{ x \pm \sqrt{x^{2} + 4x^{2}lnx +kx^{2}} }{ 2 }\]
Where k = -4C