Question

Find all solutions of xy'=2-x+(2x-2)y-xy^2.

Answer 1

xy'=2-x+2xy-2y-xy^2
xy'=-(xy^2-2xy+x)+2(1-y)
xy'=-x(y-1)^2-2(y-1)
y'=-(y-1)^2-2(y-1)/x
u=y-1
u'=y'
u'=-u^2-2u/x
w=ux
What should I do now? w=ux is the substitution that should give me a separable DE.

Answer 2

@hartnn @ganeshie8

Answer 3

w = ux
or
w= u/x
?

Answer 4

I need to find w' but what's w'? I know that it's the derivative of w and I need to use the product rule but what's w'?

Answer 5

w=ux

Answer 6

w= ux

product rule

w' = u x' + u'x
w' = u +u'x

Answer 7

How did you get w'=u+u'x from w'=ux'+u'x?

Answer 8

y = f(x)
u = y-1 = f(x) - 1
w = ux = function of x too

Answer 9

w=ux
product rule
\((fg)' = f'g +fg'\)
\((ux)' = u'x+ux'\)
x'=1

Answer 10

the independent variable is x in all your substitutions

Answer 11

you're assuming implicitly :

\(y' = \dfrac{dy}{d\color{Red}{x}}\)

\(u' = \dfrac{du}{d\color{Red}{x}}\)

\(w' = \dfrac{dw}{d\color{Red}{x}}\)

Answer 12

Now I understand how w'=u+u'x but what do I do next? I need to get a separable DE. I'm so sorry for the slowness of my internet, @hartnn @ganeshie8

Answer 13

u'=-u^2-2u/x

i would substitute u = wx

Answer 14

then you get : u' = w'x + w

Answer 15

w=u/x work ?
in my attempt, it didn't...

Answer 16

Oh the given equation is not homogeneous to start with

Answer 17

it wont work yeah

Answer 18

right,
we need to identify the form of DE ..

Answer 19

first order non-linear

Answer 20

So does the substitution u=wx and u'=w+w'x works?

Answer 21

w=ux was given as a part of hint or something or its your try ?

Answer 22

Part of hint.

Answer 23

u' = (xw' -w)/x^2

Answer 24

= -w^2 /x^2 -2w/x^2

xw' -w = -w^2 -2w

too much algebra

Answer 25

but finally , magically, its in the separable form!

Answer 26

so yes
w=ux
worked

Answer 27

could have never guessed it without the hint :P

Answer 28

So I got xw'=-w(w+1)

Answer 29

yep

Answer 30

x dw/dx = -w (w+1)

dx/x = - dw/ w(w+1)

quite simple

Answer 31

did u get this :
u' = (xw' -w)/x^2

Answer 32

Yes. :)

Answer 33

nice :)

Answer 34

then you should not have any problems solving it further :)

Answer 35

Did you know that I've struggled with this problem since I got u'=(xw'-wx')/x^2 instead of u'=(xw'-w)/x^2? HAHAHA, I'm so stupid. Thank you so much, @hartnn !!

Answer 36

lol
happens :)