# Question regarding Bernoulli Differential Equation

1. Apr 8, 2013

### mathnoobie

In a video I was watching regarding how to solve these, the lecturer said that
the form of a Bernoulli Differential Equation is y'+P(x)y=q(x)y^n
where n>1

This means that if n = 1, it wouldn't be a Bernoulli differential equation and would be a first order linear differential equation, but if n=1, y'+P(x)y=q(x)y doesn't take the form of y'+P(x)y=q(x), so how is this a first order linear differential equation?

, here is the video in case anyone wants to check it out.

Last edited by a moderator: Sep 25, 2014
2. Apr 8, 2013

### qbert

y'+P(x)y=q(x)y
y' + (P-q)y = 0

so P is P-q, and q is 0.

3. Apr 8, 2013

### mathnoobie

I don't quite understand what you did, if n=1

y'(y^-1)+P(x)-q(x)=0

4. Apr 8, 2013

### qbert

standard form for a linear first order ode
y' + f(x) y = g(x)

a n=1 example of a Bernoulli like eqn
is y' + P(x) y = q(x) y
rewrite it as y' + (P(x) - q(x))y = 0.
this is in standard form with f(x) = P(x)-q(x) and g(x) = 0.

5. Apr 8, 2013

### mathnoobie

Ah, but my original question was whether this was a Bernoulli equation or a First Order Linear DFQ(if it is, why), b/c in the video, they stated n=1 as a First Order LinDFQ

6. Apr 8, 2013

### LCKurtz

Do you not understand that the above posts are explaining to you that it is a first order linear DE and why?

7. Apr 8, 2013

### mathnoobie

Sorry, I don't. I'm not fluent enough to understand his proof without further explanation.

edit:AH, I think I finally understand the proof.

This means that the integrating factor would be e^(∫(fx)dx) correct?

Last edited: Apr 8, 2013