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

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

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

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

zepdrix
 one year ago
Best 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 \(\large u'=y^{2}\)

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

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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Haven't seen bernoulli in some time.

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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0u' = y^2y' is what your going to substitute. essentially, you're going to get a linear differential euqation.

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0You should get something like: u' + p(x)u=g(x)

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0after you substitute everything.

zepdrix
 one year ago
Best 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 :)

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

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

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

zepdrix
 one year ago
Best 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 \(\large u\).

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

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3I assume we're taking the derivative with respect to \(\large x\). That's probably why we get a prime term on each side :d

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Yes, 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.

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

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

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

zepdrix
 one year ago
Best 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}

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

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

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

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Yay 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! :)

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Oh 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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