# Find all solutions?

1. Oct 30, 2014

### Math10

1. The problem statement, all variables and given/known data
Find all solutions of xy'=2-x+(2x-2)y-xy^2.

2. Relevant equations
None.

3. 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
And now I'm stucked. Please help me, I've struggled with this problem for a long time.

2. Oct 30, 2014

### Staff: Mentor

You lost a minus sign. In the equation before the one above, you have -xy2 that became xy2.
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."

3. Oct 30, 2014

### HallsofIvy

Staff Emeritus
You lost more than just the negative on "$-xy^2$"! You should have
$$xy'= -(xy^2- 2yx+ x)+ 2(1- y)$$

4. Oct 30, 2014

### Staff: 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.

5. Oct 30, 2014

### Math10

Let me try.

6. Oct 30, 2014

### Math10

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.

7. Oct 30, 2014

### haruspex

You didn't mean that, right?

8. Oct 31, 2014

### Math10

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

9. Oct 31, 2014

### Staff: Mentor

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?

10. Oct 31, 2014

### Math10

Let me try.

11. Oct 31, 2014

### Math10

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?

12. Oct 31, 2014

### haruspex

and x' equals what?
There is a fairly obvious cancellation.

13. Oct 31, 2014

### Math10

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?

14. Oct 31, 2014

### haruspex

I ask again, what is x' equal to? Think about what it means.
There's no point in reintroducing u.

15. Oct 31, 2014

### Math10

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

16. Oct 31, 2014

### Staff: Mentor

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?

17. Oct 31, 2014

### Staff: Mentor

No.
What does x' mean?

18. Oct 31, 2014

### RUber

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

19. Nov 1, 2014

### Math10

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

20. Nov 1, 2014

### Staff: Mentor

What is x'?

By that, I mean what does x' mean?

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