# Find all solutions?

## Homework Statement

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

None.

## The Attempt at a Solution

The answer in the book is y=1-1/(x(1-cx)).
Here's my work:
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

Mark44
Mentor

## Homework Statement

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

None.

## The Attempt at a Solution

The answer in the book is y=1-1/(x(1-cx)).
Here's my work:
xy'=2-x+2xy-2y-xy^2
xy'=(xy^2-2xy+x)+2(1-y)
You lost a minus sign. In the equation before the one above, you have -xy2 that became xy2.
Math10 said:
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
I don't have any more suggestions at the moment, but I'll take a closer look in a little while.

BTW, there is no such word in English as "stucked."

HallsofIvy
Homework Helper

## Homework Statement

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

None.

## The Attempt at a Solution

The answer in the book is y=1-1/(x(1-cx)).
Here's my work:
xy'=2-x+2xy-2y-xy^2
xy'=(xy^2-2xy+x)+2(1-y)
You lost more than just the negative on "$-xy^2$"! You should have
$$xy'= -(xy^2- 2yx+ x)+ 2(1- y)$$

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

Mark44
Mentor
Math10,
After you fix your algebra errors, with your substitution you should have an equation in x, u, and u'. Try the substitution w = ux. That should get you a new DE that is separable.

Let me try.

Here's my work:

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
u=w/x
u'=w'
w'=-(w^2/x^2)-(2w/x)(1/x)
w'=-w^2/x^2-2w/x^2
w'=(-w^2-2w)/x^2
dw/(-w^2-2w)=dx/x^2
dw/(-w(w+2))=dx/x^2
A/-w+B/(w+2)=1
A(w+2)-Bw=1
A=1/2, B=1/2
(-1/2)ln abs(w)+(1/2)ln abs(w+2)=-1/x+C
(-1/2)(ln abs(w)-ln abs(w+2))=-1/x+C
ln abs(w/(w+2))=2/x+C
w/(w+2)=Ce^(2/x)
w=Ce^(2/x)(w+2)
w=wCe^(2/x)+2Ce^(2/x)
w-wCe^(2/x)=2Ce^(2/x)
w(1-Ce^(2/x))=2Ce^(2/x)
w=2Ce^(2/x)/(1-Ce^(2/x))
u=2Ce^(2/x)/(x(1-Ce^(2/x)))
y=1+2Ce^(2/x)/(x(1-Ce^(2/x)))
This doesn't match the answer in my book. Please check my work carefully and tell me where I got wrong if I'm incorrect.

haruspex
Homework Helper
Gold Member
2020 Award
u=w/x
u'=w'
You didn't mean that, right?

So where did I make a mistake? What's wrong?

Mark44
Mentor
u=w/x
u'=w'
haruspex said:
You didn't mean that, right?
So where did I make a mistake? What's wrong?
You're treating x as if it were a constant - it's not. Instead of working with the equation, use the equivalent equation w = vx. Now, what is w'? What differentiation rule should come to mind?

Let me try.

I see how I got it wrong.
So with the substitution w=ux,
u=w/x
u'=(xw'-wx')/x^2
(xw'-wx')/x^2=-(w^2/x^2)-(2w/x)(1/x)
(xw'-wx')/x^2=(-w^2-2w)/x^2
Now what should I do?

haruspex
Homework Helper
Gold Member
2020 Award
I see how I got it wrong.
So with the substitution w=ux,
u=w/x
u'=(xw'-wx')/x^2
and x' equals what?
(xw'-wx')/x^2=(-w^2-2w)/x^2
Now what should I do?
There is a fairly obvious cancellation.

So I got
xw'-wx'=-w^2-2w
wx'=xw'+w^2+2w
x'=(xw'+w^2+2w)/w
x'=w'/u+w+2
Now what?

haruspex
Homework Helper
Gold Member
2020 Award
So I got
xw'-wx'=-w^2-2w
I ask again, what is x' equal to? Think about what it means.
x'=w'/u+w+2
There's no point in reintroducing u.

x'=w'x/w+w+2
Isn't it?

Mark44
Mentor
I see how I got it wrong.
So with the substitution w=ux,
Why did you make it harder on yourself by solving for u again? It's a lot simpler to differentiate the equation above than the one you have below.

And think about what you're doing. You're differentiating with respect to which variable?
Math10 said:
u=w/x
u'=(xw'-wx')/x^2
(xw'-wx')/x^2=-(w^2/x^2)-(2w/x)(1/x)
(xw'-wx')/x^2=(-w^2-2w)/x^2
Now what should I do?

Mark44
Mentor
x'=w'x/w+w+2
Isn't it?
No.
What does x' mean?

RUber
Homework Helper
The assumption for the beginning was that y was a function of x.
y'=dy/dx.

So w'=ux'+xu' and what do I do next?

Mark44
Mentor
So w'=ux'+xu' and what do I do next?
What is x'?

By that, I mean what does x' mean?

Mark44
Mentor
Here's my work:

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
u=w/x
u'=w'
w'=-(w^2/x^2)-(2w/x)(1/x)
w'=-w^2/x^2-2w/x^2
w'=(-w^2-2w)/x^2
dw/(-w^2-2w)=dx/x^2
dw/(-w(w+2))=dx/x^2
A/-w+B/(w+2)=1
A(w+2)-Bw=1
A=1/2, B=1/2
(-1/2)ln abs(w)+(1/2)ln abs(w+2)=-1/x+C
(-1/2)(ln abs(w)-ln abs(w+2))=-1/x+C
ln abs(w/(w+2))=2/x+C
w/(w+2)=Ce^(2/x)
w=Ce^(2/x)(w+2)
w=wCe^(2/x)+2Ce^(2/x)
w-wCe^(2/x)=2Ce^(2/x)
w(1-Ce^(2/x))=2Ce^(2/x)
w=2Ce^(2/x)/(1-Ce^(2/x))
u=2Ce^(2/x)/(x(1-Ce^(2/x)))
y=1+2Ce^(2/x)/(x(1-Ce^(2/x)))
This doesn't match the answer in my book. Please check my work carefully and tell me where I got wrong if I'm incorrect.
Your work above is what we call a "wall of text." I don't see a single space anywhere in your equations. This makes it much more difficult to understand what you have written. You can make things more readable by inserting spaces around addition and subtraction operations, and around '='.