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dfreseniusBest ResponseYou've already chosen the best response.0
Yes, this is a bit strange because they want me to substitute u=y^(1n)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
so, im thinking to start off, i will get x and y on one side.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362709077091:dw
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
i suppose i should state what bernoulli is
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
\[\large xy' + y = −2xy^2\]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 \(\large u'=y^{2}\)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Woops, \(\large u'=y^{2}y'\)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362709205198:dw
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
Haven't seen bernoulli in some time.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ya im a bit shaky on this but im caught up to you zep, so the u'=y'y^2 , can you explain that?
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
u' = y^2y' is what your going to substitute. essentially, you're going to get a linear differential euqation.
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
You should get something like: u' + p(x)u=g(x)
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
after you substitute everything.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
\[\large y'+p(x)y=q(x)y^n\] Dividing both sides by y^n, \(\large \color{royalblue}{y^{n}y'}+p(x)\color{orangered}{y^{1n}}=q(x)\) Making the sub \(\large \color{orangered}{u=y^{1n}}\) \(\large u'=(1n)y^{n}y' \qquad \rightarrow \qquad \color{royalblue}{\dfrac{1}{1n}u'=y^{n}y'}\) Giving us, yah something like what abbot said :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
I tried to use colors to make that clearer, but it's still rather confusing lol
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ya, im just puzzled
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
so, with my equation,
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
\[\large xy' + y = −2xy^2\]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,\[\large y^{2}y' + \frac{1}{x}\color{royalblue}{y^{12}} = −2\]Then we want to make that blue term our \(\large u\).
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362709954712:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Our \(\large n\) in this case is 2, because our y^n was the term on the RIGHT side of the equation. \(\large u=y^{1n} \qquad \rightarrow \qquad u=y^{21}\qquad \rightarrow \qquad u=y^{1}\)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ok that makes sense
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362710284245:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
I think maybe you missed a negative when taking the derivative of your u. \(\huge u=y^{1} \qquad \rightarrow \qquad u'=\color{red}{}y^{2}y'\)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
oh the derivative gotcha
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
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
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362710497399:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Remember that before you can get an integrating factor and all that fun stuff, your leading term needs to be positive.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
should i just switch sides?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Just multiply everything by 1 :)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362710598175:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Woops, standard form looks like this,\[\large u'+P(x)u=Q(x)\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
You had it in your middle step.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362710722310:dw
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ok so, i integrate both sides?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
No you need to find an integrating factor \(\large \mu=e^{\int P(x)\;dx}\) Look familiar I hope? :)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ya let me check real quick
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362710969094:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Woops the negative sign is also part of your \(\large P(x)\). Recall that the standard form shows addition in the middle there.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dangit, i need to stop rushing
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
And your equation should be \(\large u'\dfrac{1}{x}u=2\) Positive 2 on the right. I think we missed that earlier.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Yes you do need to stop rushing Chris Farley!! Slow down in life! Or you'll end up leaving us too soon......... :'C
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362711281571:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
\[\huge e^{\ln x} \qquad = \qquad e^{\ln (x^{1})} \qquad = \qquad ?\]
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
ok so now that i have that, i multiply by x^1
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
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.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362711568703:dw
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
now, i want to isolate u' right?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
From here, you want to recognize that the LEFT side is the result of the product rule.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
\[\large \left[u(x)f(x)\right]' \qquad = \qquad u'(x)f(x)+u(x)f'(x)\] We do that have in this case don't we? Where our f(x)=x^{1}
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
yes, so I can simplify the left
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362711903145:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Oh sorry I forget you were calling it \(\large I\). I hope I didn't make that more confusing c:
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
yah everything looks good so far.
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362712182784:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Yes good. Then we have to do a little work changing our \(\large u\) back to \(\large y\)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Good.\[\large \frac{1}{y}=2x \ln x+Cx\]So how do we solve for y?
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
dw:1362712446678:dw
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
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%3D2xy%5E2 Same result! Yay! :)
 one year ago

dfreseniusBest ResponseYou've already chosen the best response.0
thank you so much for your patience, i have been doing very well in my class because of your help
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Oh that's good to hear! c: I'm sure it's not just me. You seem to understand these concepts very very quickly.
 one year ago
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