# Is this a Bernoulli diff.eqn?

1. Sep 20, 2010

### Susanne217

1. The problem statement, all variables and given/known data

This diff.eqn here.

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

which in my mind can be re-written to

$$\frac{dy}{dx} -\frac{1}{x}y = -y^2$$

does this mean that the above is a socalled bernoulli diff.eqn? And should solved as such? I may be dumb, but that the only way I can deduce is to solve that eqn..

Sincerely
Susanne

2. Sep 20, 2010

### LawlQuals

Sure, looks Bernoulliriffic to me. Just trust yourself, go through the solution strategy for the Bernoulli equations, and you will arrive at a result. It is always good to try things out before asking questions. Maybe you can solve that with exact differentials + integrating factor as well, but at least using solution methods a la Bernoulli works out just fine too.

3. Sep 20, 2010

### Susanne217

Okay thanks Lawl,

The reason I'm asking is that I get the right solution using Maple and the wrong one by hand :(

But now I know that it is a Bernoulli diff.eqn thanks :)

But why do I keep getting the wrong result by hand??:uhh:

Maple says to me Susanne the solution is

$$y(x) = \frac{2x}{x^2+2\cdotC}$$

But then I use my own brain I get a totally different result. Look.

First I choose $$w = y^{-1}$$

Why by the solution method for Bernoulli diff.eqn from my textbook( Zigs First Course in differential eqn p. 62-63).

$$\frac{dw}{dx}+ \frac{1}{x}w = 1$$

I find the integration factor to be x^-1.

thus I get that $$x^{-1} \cdot w = \int(\frac{1}{x})dx = ln(x)$$

and by replacing w with y^-1

I get

$$y(x) = \frac{1}{x\cdot ln(x)+kx}$$

Which as you can see is lightyears away from what Maple says the solution. What am I doing wrong??? Please point out where I in my calc are doing wrong :(

Last edited: Sep 20, 2010
4. Sep 20, 2010

### LawlQuals

The incongruence between your work and the solution quoted by Maple stems from the integrating factor. Please examine that again. And, note that Maple has (for some reason) chosen the constant of integration in your solution to be equal to unity.