Differential equation w/o an x variable

[robbondo]

Homework Statement

y\prime = ay - by^{2}

Homework Equations

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.

 

[dashkin111]

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

 

[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?

 

[dashkin111]

No

Try factoring more like this:

<br /> ay-by^{2}=y(a-by)<br />

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

 

[robbondo]

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

 

[dashkin111]

Sooo...

<br /> \int\frac{dy}{y(a-yb)}=\int dx<br />

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

 

[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.

 

[dashkin111]

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.

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

 

[Dick]

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

 

[dashkin111]

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

 

[robbondo]

So, when I did the separation 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? :)

 

[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.

 

[robbondo]

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

 

[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}}

 

[Dick]

That looks much better.

 

[robbondo]

Thanks!