# O.D.E. question too

1. Aug 27, 2005

### asdf1

In this O.D.E. question,
xy=y^2 +y
i don't think it's separable,
but i'm not sure what other steps to take...
does someone have an idea?
btw, what do most people need to think of when they encounter O.D.E. problems?

2. Aug 27, 2005

### lurflurf

It is obviously seperatable
y'/(y^2+y)=1/x
it may be helpful to write it in the form
(y^-2)y'/(1+y^-1)=1/x
I do not know what most people think when they enconter O.D.E.
I know what I think when I see this one
-This is obviously separatable
-I see logarithmic derivatives (things of form f'/f)

3. Aug 29, 2005

### asdf1

:P
sorry, my mistake!
hmm... so
dy/(y^2+y)=xdx
but how do you integrate the left side?

4. Aug 29, 2005

### TD

I think you mean:

$$\frac{{dy}}{{y^2 + y}} = \frac{{dx}}{x}$$

You can either do what lurflurf said, but if you don't see that you could just split in partial fractions, giving simple logarithms:

$$\frac{1}{{y^2 + y}} = \frac{1}{y} - \frac{1}{{y + 1}}$$

5. Aug 29, 2005

### asdf1

thank you very much!!! :)
if you want to do it using lurflurf's way, using logarithmic derivatives, how is that done?

6. Aug 29, 2005

### TD

This one will give logarithmes as well - since we have f'/f too - it's just that lurflurf re-wrote it in such a way that the partial fractions weren't necessary.
When you don't just "see" that like lurflurf, I suggest you do it like above.

7. Aug 30, 2005

### asdf1

ok, good suggestion~
thanks again!!!

8. Sep 3, 2005

### asdf1

ok, here's my work:
dy/(1+y^2 +y)=dx/x
=> [1/y-1/(y+1)]=dx/x
=> ln[absolute value (y)]- ln{absolute value [1/(y+1)]}=ln[absolute value (x)]+c
=> ln{absolute value [y/(y+1)]}=ln[absolute value (x)]+c`
=> y/(y+1)=cx
however, the correct answer should be y=x/(c-x)
does anybody know where my calculations went wrong?

9. Sep 3, 2005

### lurflurf

Those are the same, one is in terms of x and one y and their constants are multiplicative inverses, but they represent the same family of solutions.

10. Sep 3, 2005

### asdf1

@@a
that idea is new to me, so i'm still in a little surprised state~
you mean you can flip-flop the x and y variables?
but i thought that you can start out with same x and y variable in the same state, so it shouldn't be alright to flip-flop them? (i mean if you double-checked that O.D.E by using the v=y/x substitution, you start out at the same step, because that ends up with the original answer...)

11. Sep 3, 2005

### lurflurf

I don't mean interchange them. I mean solving either equation for the other variable gives the other equation. The constants are multiplicative inverses so I will use c and c' to represent them.
y/(y+1)=cx
(y+1)/y=1/(cx)
1+1/y=1/(cx)
1/y=1/(cx)-1
y=1/(1/(cx)-1)
y=cx/(1-cx)
y=x/(1/c-x)
y=x/(c'-x)
ditto reverse
y=x/(c'-x)
y=x/(1/c-x)
y=cx/(1-cx)
y=1/(1/(cx)-1)
1/y=1/(cx)-1
1+1/y=1/(cx)
(y+1)/y=1/(cx)
y/(y+1)=cx

12. Sep 4, 2005

### asdf1

i see now...
thank you very much!!! :)

13. Sep 4, 2005

### asdf1

by the way, that's amazing!
how'd you think of doing it that way?

14. Sep 4, 2005

### Galileo

He just solved y/(y+1)=cx for y, which is obviously what you're after.
If you solve it you get:

$$y=\frac{cx}{1-cx}=\frac{x}{1/c-x}$$
Now 1/c is just as arbitrary a variable as c, so it doesn't matter which one you use. You could introduce a new variable c'=1/c and you have the solution in the same form as the book.

15. Sep 4, 2005

### lurflurf

It is very common in working a differential equation to find answers that very in form. It is traditional to consider the variable differented (y in the question as asked) to be the dependent variable and express the answer as that variable as function a function of the other when it is easily done. When confronted with two answer that are not immediately seen to be equivalent there are a few things that can be done
-If it is know that each has the form of the most general solution (with possible exceptions for singular solutions) checking that each satisfys the original question is sufficient.
-one may do as I have done and reduce one two the form of the other (it is possible that one form is more general than the other if some steps in such a reduction are not reversible)
-substitute one equation into the other and show an identity results.

In this particular case I saw by your work that you were likely correct (My answer mached the other form). I then made doubly sure my checking your equation in the original equation. Convinced that you were correct I set about reducing each equation to the other, as I knew getting the same answer in different forms is a common occurance.

16. Sep 4, 2005

### asdf1

wow~ thank you very much!!! :)