Solve the Bernoulli equation.

**A Beautiful Mind** · September 22nd 2010, 10:17 PM

$\frac{dy}{dx}+xy=xy^3$

I have no idea how to do these. And we *have* to do it as a Bernoulli equation.

Initial values: (0, 1/3)

**mr fantastic** · September 22nd 2010, 10:37 PM

Originally Posted by A Beautiful Mind

$\frac{dy}{dx}+xy=xy^3$

I have no idea how to do these. And we *have* to do it as a Bernoulli equation.

Initial values: (1/3, 0)

The procedure is given here (and no doubt it's in your classnotes and textbook too): Bernoulli differential equation - Wikipedia, the free encyclopedia

What part of applying this procedure are you stuck on?

**A Beautiful Mind** · September 22nd 2010, 11:02 PM

Well, my answer I get before using the initial conditions is:

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

$w'+xw=x$ , where $w = \frac{1}{y^2}$
$w' = -3y^{-4}$
$w'=\frac{-3}{y^{4}}$

$e^{\frac{x^2}{2}$ is the integrating factor...the rest I tried to put on here, but the stuff keeps getting messed up in the latex.

**mr fantastic** · September 22nd 2010, 11:10 PM

Originally Posted by A Beautiful Mind

Well, my answer I get before using the initial conditions is:

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

http://www.wolframalpha.com/input/?i...+xy+%3D+xy%5E3

The DE is seperable in fact. Are you required to do it as a Bernoulli?

**A Beautiful Mind** · September 22nd 2010, 11:14 PM

Yes, we're required to do it as a Bernoulli. I noticed it was separable too, but she's very strict about this.

**A Beautiful Mind** · September 24th 2010, 12:55 PM

I still need help with this...I'm getting nowhere. And my teachers office hours are cancelled. And the tutors can't do above Calc III. :/

**mr fantastic** · September 24th 2010, 02:17 PM

Originally Posted by A Beautiful Mind

Well, my answer I get before using the initial conditions is:

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

$w'+xw=x$ , Mr F says: This is wrong.

where $w = \frac{1}{y^2}$ Mr F says: Yes, this is the correct substitution.

$w' = -3y^{-4}$ Mr F says: This is wrong.

$w'=\frac{-3}{y^{4}}$

$e^{\frac{x^2}{2}$ is the integrating factor...the rest I tried to put on here, but the stuff keeps getting messed up in the latex.

There are a number of things you are not doing correctly. Have you read any references on the required technique eg. http://en.wikipedia.org/wiki/Bernoul...ntial_equation.

Please show all your work, every step, so that it can be reviewed.

**A Beautiful Mind** · September 24th 2010, 02:40 PM

$\frac{dy}{dx}+xy=xy^3$

$n = 3$

$u = y^{1-3}$

$u = y^-2<br />$
$\frac{du}{dx}$

$\frac{du}{dx} = -2y^{-3}$

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

$\frac{1}{-2y^{-3}}\frac{du}{dx}+xy=xy^3$

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

Getting integrating factor.

$P(x) = -2x$

$e^{\int-2x} = e^{-x^2}$

$e^{-x^2}\frac{du}{dx}+(-2y^{-2}xe^{-x^2}) = -2xe^{-x^2}$

Did the integration in my head. You integrate [e^-x^/2(u)}]' and -2xe^{-x^2}. It's easy to see on the right side the -2x term will cancel out by u-substitution.

$e^{-x^2}u = e^-x^2+C$

Sub in the u.

$\frac{e^{-x^2}}{y^2} = e^-x^2+C$

$\frac{1}{y^2}=1+Ce^{x^2}$

Plug in initial values to this equation. Y(0) = 1/3

$\frac{1}{(1/3)^2} = 1+Ce^0$
$9 = 1+C$
$C= 8$

$\frac{1}{y^2}= 1+8e^{x^2}$

And yes, I read your first reference to wikipedia, but I got the first answer I posted yesterday as that weird square root. Then, I tried looking for examples.

**mr fantastic** · September 24th 2010, 02:55 PM

Originally Posted by A Beautiful Mind

$\frac{dy}{dx}+xy=xy^3$

$n = 3$

$u = y^{1-3}$

$u = y^-2<br />$
$\frac{du}{dx}$

$\frac{du}{dx} = -2y^{-3}$

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

$\frac{1}{-2y^{-3}}\frac{du}{dx}+xy=xy^3$ Mr F says: All OK to here (although there are parts of your working I find difficult to understand)

$\frac{du}{dx}-2y^{-2}=-2x$ Mr F says: This is wrong. There is a missing x. It should be $\displaystyle \frac{du}{dx} - 2xy^{-2} = -2x$ . From which you get $\displaystyle \frac{du}{dx} - 2ux = -2x$ .

Getting integrating factor.

$P(x) = -2x$

$e^{\int-2x} = e^{-x^2}$

$e^{-x^2}\frac{du}{dx}+(-2y^{-2}xe^{-x^2}) = -2xe^{-x^2}$

Did the integration in my head. You integrate [e^-x^/2(u)}]' and -2xe^{-x^2}. It's easy to see on the right side the -2x term will cancel out by u-substitution.

$e^{-x^2}u = e^-x^2+C$

Sub in the u.

$\frac{e^{-x^2}}{y^2} = e^-x^2+C$

$\frac{1}{y^2}=1+Ce^{x^2}$

Plug in initial values to this equation. Y(0) = 1/3

$\frac{1}{(1/3)^2} = 1+Ce^0$
$9 = 1+C$
$C= 8$

$\frac{1}{y^2}= 1+8e^{x^2}$

And yes, I read your first reference to wikipedia, but I got the first answer I posted yesterday as that weird square root. Then, I tried looking for examples.

OK, this final answer is correct. But a lot of your working presents as wrong even though you finally get the correct answer. You will lose marks for that.

**A Beautiful Mind** · September 24th 2010, 02:58 PM

Sorry, the x reappears in the next step, though. I subbed u =y^-2 though in that part.

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