$\displaystyle \frac{dy}{dx}+xy=xy^3$
I have no idea how to do these. And we *have* to do it as a Bernoulli equation.
Initial values: (0, 1/3)
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?
Well, my answer I get before using the initial conditions is:
$\displaystyle y = \sqrt{1+Ce^\frac{-x^2}{2}}$
$\displaystyle w'+xw=x$, where $\displaystyle w = \frac{1}{y^2}$
$\displaystyle w' = -3y^{-4}$
$\displaystyle w'=\frac{-3}{y^{4}}$
$\displaystyle e^{\frac{x^2}{2}$ is the integrating factor...the rest I tried to put on here, but the stuff keeps getting messed up in the latex.
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?
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.
$\displaystyle \frac{dy}{dx}+xy=xy^3$
$\displaystyle n = 3$
$\displaystyle u = y^{1-3}$
$\displaystyle u = y^-2
$
$\displaystyle \frac{du}{dx}$
$\displaystyle \frac{du}{dx} = -2y^{-3}$
$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
$\displaystyle \frac{1}{-2y^{-3}}\frac{du}{dx}+xy=xy^3$
$\displaystyle \frac{du}{dx}-2y^{-2}=-2x$
Getting integrating factor.
$\displaystyle P(x) = -2x$
$\displaystyle e^{\int-2x} = e^{-x^2}$
$\displaystyle e^{-x^2}\frac{du}{dx}+(-2y^{-2}xe^{-x^2}) = -2xe^{-x^2}$
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.
$\displaystyle e^{-x^2}u = e^-x^2+C$
Sub in the u.
$\displaystyle \frac{e^{-x^2}}{y^2} = e^-x^2+C$
$\displaystyle \frac{1}{y^2}=1+Ce^{x^2}$
Plug in initial values to this equation. Y(0) = 1/3
$\displaystyle \frac{1}{(1/3)^2} = 1+Ce^0$
$\displaystyle 9 = 1+C$
$\displaystyle C= 8$
$\displaystyle \frac{1}{y^2}= 1+8e^{x^2}$
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.