Question

Solve (y'x-y)(ln abs(y)-ln abs(x))=x implicitly.

Answer 1

y=ux
u=y/x
y'=u+u'x
(x(u+u'x)-ux)(ln abs(ux)-ln abs(x))=x
(ux+u'x^2-ux)(ln abs(u))=x
u'x^2*ln abs(u)=x
u'x*ln abs(u)=1
u'x=1/ln abs(u)
ln abs(u) du=dx/x
integration by parts
t=ln abs(u)
dv=du
dt=1/u du
v=u
u*ln abs(u)-u+C
u*ln abs(u)-u=ln abs(x)+C
(y/x)ln abs(y/x)-y/x=ln abs(x)+C

Answer 2

The answer is (x+y)ln abs(x)+y(1-ln abs(y))+Cx=0 in the book.

Answer 3

@zepdrix @phi @amistre64

Answer 4

How do I get to the answer that I provided from there?

Answer 5

multiply thru by x and factor as needed?

Answer 6

(y/x)ln abs(y/x)-y/x=ln abs(x)+C

y ln abs(y/x) -y = x ln abs(x) + Cx

y ln abs(y/x) - x ln abs(x) - y + Cx = 0

y [ln abs(y) - ln abs(x)] - x ln abs(x) - y + Cx = 0

Answer 7

y ln abs(y) - y - y ln abs(x) - x ln abs(x) + Cx = 0

y [ ln abs(y) - 1] - (x+y) ln abs(x) + Cx = 0

-y [ 1 - ln abs(y)] - (x+y) ln abs(x) + Cx = 0

Answer 8

then multiply each side by -1

Answer 9

I got the right answer! Thank you!