# Differential equation w/o an x variable

1. Oct 8, 2007

### robbondo

1. The problem statement, all variables and given/known data
$$y\prime = ay - by^{2}$$

2. Relevant equations

3. The attempt at a solution

Is this a linear DOE? If so when I use the integrating factor method, would the int. factor be
$$e^{-a\int?}$$ ? would it by dy or dx? I think I'm missing the big picture with this process.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 8, 2007

### dashkin111

Have you tried factoring out the y, dividing over, and partial fractions?

3. Oct 8, 2007

### robbondo

Hmm... Well when I factor out y and divide I get

$$\frac{y\prime}{y} = a - by$$?

can I integrate with that y on the other side still?

4. Oct 8, 2007

### dashkin111

No

Try factoring more like this:

$$ay-by^{2}=y(a-by)$$

Divide by the whole thing on the right and try partial fractions

5. Oct 8, 2007

### robbondo

I don't see how that will work... I'm confused.

6. Oct 8, 2007

### dashkin111

Sooo...

$$\int\frac{dy}{y(a-yb)}=\int dx$$

Do you understand how to do the left side? Use partial fractions

7. Oct 8, 2007

### robbondo

OK so I did the partial fractions took the integral and then did some moving around of logs to get $$y = \frac{a^{2}}{e^{x} + ab}$$ Does that look close to anything you might have gotten. I'm not super confident in how I manipulated the logs.

8. Oct 8, 2007

### dashkin111

Maybe, you should get an e somewhere in your answer. If I have time later maybe I'll check it for you.

9. Oct 8, 2007

### Dick

Nope, doesn't look quite right. And you're missing a constant of integration. Can you show more work?

10. Oct 8, 2007

### dashkin111

If you even show your partial fractions that you created, it'd be very helpful

11. Oct 8, 2007

### robbondo

So, when I did the seperation of variables I got
$$\frac{1}{a} \int{\frac{dy}{y}} + \frac{b}{a} \int{\frac{dy}{a-by}} = \int{dx}$$

then when I took the integrals I get

$$\ln{y^{-a}} - \ln{(a-by)^{-a}} = x + c$$

so then I messed around and combined the logs into a fraction, and since they were both to the -a I just switched to put the a-by on top and the a y on the bottom, then i took the whole thing to the power of e to get rid of the ln, and then brought the a out to the front.

$$a(\frac{a-by}{y}) = e^{x+c}$$ then multiplying a into the fraction and solvinf for y I get

$$\frac{a^{2}}{e^{x+c} + ab } = y$$

where'd I screw up? :)

12. Oct 8, 2007

### Dick

Put the a over on the dx side first. And I don't know how it turned into a '-a'. That 'a' should wind up inside the exponential.

13. Oct 8, 2007

### robbondo

I didn't mean separation of variable, I mean partial fractions... sorry

14. Oct 8, 2007

### robbondo

yeah multiplying that a makes it much simpler...

So then I get

$$\int{\frac{dy}{y}} + b\int{\frac{dy}{a-by}} = a \int{dx}$$

then I get

$$\ln\{\frac{y}{a-by}} = ax + c$$

then raised to the e

$$\frac{y}{a-by} = ce^{ax}$$

so then when i solve for y I get $$y = \frac{ace^{ax}}{1+bce^{ac}}$$

15. Oct 8, 2007

### Dick

That looks much better.

16. Oct 8, 2007

Thanks!