Find all solutions?

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Homework Statement


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

Homework Equations


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

Answers and Replies

  • #2
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Homework Statement


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

Homework Equations


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


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

Homework Equations


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 "[itex]-xy^2[/itex]"! You should have
[tex]xy'= -(xy^2- 2yx+ x)+ 2(1- y)[/tex]

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.
 
  • #4
35,202
7,005
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.
 
  • #6
301
0
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.
 
  • #8
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0
So where did I make a mistake? What's wrong?
 
  • #9
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7,005
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?
 
  • #11
301
0
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
haruspex
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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.
 
  • #13
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0
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
haruspex
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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.
 
  • #15
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x'=w'x/w+w+2
Isn't it?
 
  • #16
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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?
 
  • #18
RUber
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The assumption for the beginning was that y was a function of x.
y'=dy/dx.
 
  • #19
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0
So w'=ux'+xu' and what do I do next?
 
  • #21
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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 '='.

Here's one of your equation with some spaces added:
w = wCe^(2/x) + 2Ce^(2/x)
If you want us to help you, put some extra effort into making it easier for us to do so.
 

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