anonymous
  • anonymous
xy' + y = −2xy^2 Bernoulli Differential equation.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Yes, this is a bit strange because they want me to substitute u=y^(1-n)
anonymous
  • anonymous
so, im thinking to start off, i will get x and y on one side.
anonymous
  • anonymous
|dw:1362709077091:dw|

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anonymous
  • anonymous
i suppose i should state what bernoulli is
zepdrix
  • zepdrix
\[\large xy' + y = −2xy^2\]Hmm it's been a lil while.. I'm trying to remember how to do these. I think you want to get the equation to this form. ~Divide both sides by x, Divide both sides by y^2. \[\large y^{-2}y' + \frac{1}{x}y^{-1} = −2\] Then let \(\large u=y^{-1}\). Hmm \(\large u'=-y^{-2}\)
zepdrix
  • zepdrix
Woops, \(\large u'=-y^{-2}y'\)
anonymous
  • anonymous
|dw:1362709205198:dw|
abb0t
  • abb0t
Haven't seen bernoulli in some time.
anonymous
  • anonymous
ya im a bit shaky on this but im caught up to you zep, so the u'=y'y^-2 , can you explain that?
abb0t
  • abb0t
u' = y^-2y' is what your going to substitute. essentially, you're going to get a linear differential euqation.
abb0t
  • abb0t
You should get something like: u' + p(x)u=g(x)
abb0t
  • abb0t
after you substitute everything.
zepdrix
  • zepdrix
\[\large y'+p(x)y=q(x)y^n\] Dividing both sides by y^n, \(\large \color{royalblue}{y^{-n}y'}+p(x)\color{orangered}{y^{1-n}}=q(x)\) Making the sub \(\large \color{orangered}{u=y^{1-n}}\) \(\large u'=(1-n)y^{-n}y' \qquad \rightarrow \qquad \color{royalblue}{\dfrac{1}{1-n}u'=y^{-n}y'}\) Giving us, yah something like what abbot said :)
zepdrix
  • zepdrix
I tried to use colors to make that clearer, but it's still rather confusing lol
anonymous
  • anonymous
ya, im just puzzled
anonymous
  • anonymous
so, with my equation,
zepdrix
  • zepdrix
\[\large xy' + y = −2xy^2\]Dividing each side by x gives us,\[\large y' + \frac{1}{x}y = −2y^2\]Dividing each side by y^2 (Because that is our y^n is this case) gives,\[\large y^{-2}y' + \frac{1}{x}\color{royalblue}{y^{1-2}} = −2\]Then we want to make that blue term our \(\large u\).
anonymous
  • anonymous
|dw:1362709954712:dw|
zepdrix
  • zepdrix
Our \(\large n\) in this case is 2, because our y^n was the term on the RIGHT side of the equation. \(\large u=y^{1-n} \qquad \rightarrow \qquad u=y^{2-1}\qquad \rightarrow \qquad u=y^{-1}\)
anonymous
  • anonymous
ok that makes sense
anonymous
  • anonymous
|dw:1362710284245:dw|
zepdrix
  • zepdrix
I think maybe you missed a negative when taking the derivative of your u. \(\huge u=y^{-1} \qquad \rightarrow \qquad u'=\color{red}{-}y^{-2}y'\)
anonymous
  • anonymous
oh the derivative gotcha
zepdrix
  • zepdrix
I assume we're taking the derivative with respect to \(\large x\). That's probably why we get a prime term on each side :d
anonymous
  • anonymous
|dw:1362710497399:dw|
zepdrix
  • zepdrix
Yes good! :)
zepdrix
  • zepdrix
Remember that before you can get an integrating factor and all that fun stuff, your leading term needs to be positive.
anonymous
  • anonymous
ok
anonymous
  • anonymous
should i just switch sides?
zepdrix
  • zepdrix
Or maybe a better way to say that would be, it needs to be in standard form* no coefficients or anything on the leading term.
zepdrix
  • zepdrix
Just multiply everything by -1 :)
anonymous
  • anonymous
|dw:1362710598175:dw|
zepdrix
  • zepdrix
Woops, standard form looks like this,\[\large u'+P(x)u=Q(x)\]
zepdrix
  • zepdrix
You had it in your middle step.
anonymous
  • anonymous
|dw:1362710722310:dw|
zepdrix
  • zepdrix
heh XD
anonymous
  • anonymous
ok so, i integrate both sides?
zepdrix
  • zepdrix
No you need to find an integrating factor \(\large \mu=e^{\int P(x)\;dx}\) Look familiar I hope? :)
anonymous
  • anonymous
ya let me check real quick
anonymous
  • anonymous
|dw:1362710969094:dw|
zepdrix
  • zepdrix
Woops the negative sign is also part of your \(\large P(x)\). Recall that the standard form shows addition in the middle there.
anonymous
  • anonymous
dangit, i need to stop rushing
zepdrix
  • zepdrix
And your equation should be \(\large u'-\dfrac{1}{x}u=2\) Positive 2 on the right. I think we missed that earlier.
zepdrix
  • zepdrix
Yes you do need to stop rushing Chris Farley!! Slow down in life! Or you'll end up leaving us too soon......... :'C
anonymous
  • anonymous
haha nice
anonymous
  • anonymous
|dw:1362711281571:dw|
zepdrix
  • zepdrix
\[\huge e^{-\ln x} \qquad = \qquad e^{\ln (x^{-1})} \qquad = \qquad ?\]
anonymous
  • anonymous
1/x
zepdrix
  • zepdrix
Cool c:
anonymous
  • anonymous
my bad
anonymous
  • anonymous
ok so now that i have that, i multiply by x^-1
zepdrix
  • zepdrix
Yes, if you want to be thorough, then multiply through by your integrating factor. I usually skip that step. But it might be good not to skip it since you're learning it still.
anonymous
  • anonymous
ok,
anonymous
  • anonymous
|dw:1362711568703:dw|
anonymous
  • anonymous
now, i want to isolate u' right?
zepdrix
  • zepdrix
Looks good so far c: BTW, just a little note: You can scale the size of the drawing box if you don't want to take up that much space.|dw:1362711683197:dw|
zepdrix
  • zepdrix
From here, you want to recognize that the LEFT side is the result of the product rule.
zepdrix
  • zepdrix
\[\large \left[u(x)f(x)\right]' \qquad = \qquad u'(x)f(x)+u(x)f'(x)\] We do that have in this case don't we? Where our f(x)=x^{-1}
anonymous
  • anonymous
yes, so I can simplify the left
zepdrix
  • zepdrix
ya sounds good c:
anonymous
  • anonymous
|dw:1362711903145:dw|
zepdrix
  • zepdrix
Oh sorry I forget you were calling it \(\large I\). I hope I didn't make that more confusing c:
zepdrix
  • zepdrix
yah everything looks good so far.
anonymous
  • anonymous
|dw:1362712182784:dw|
anonymous
  • anonymous
Cx
zepdrix
  • zepdrix
Yes good. Then we have to do a little work changing our \(\large u\) back to \(\large y\)
anonymous
  • anonymous
ok, u=y^-1
zepdrix
  • zepdrix
Good.\[\large \frac{1}{y}=2x \ln x+Cx\]So how do we solve for y?
anonymous
  • anonymous
|dw:1362712446678:dw|
zepdrix
  • zepdrix
Yay good job! I threw it into Wolfram just a moment ago to make sure we didn't make any fatal errors. http://www.wolframalpha.com/input/?i=xy%27%2By%3D-2xy%5E2 Same result! Yay! :)
anonymous
  • anonymous
YESSS
anonymous
  • anonymous
thank you so much for your patience, i have been doing very well in my class because of your help
zepdrix
  • zepdrix
Oh that's good to hear! c: I'm sure it's not just me. You seem to understand these concepts very very quickly.

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