A community for students.
Here's the question you clicked on:
 0 viewing
dfresenius
 2 years ago
xy' + y = −2xy^2 Bernoulli Differential equation.
dfresenius
 2 years ago
xy' + y = −2xy^2 Bernoulli Differential equation.

This Question is Closed

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

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

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

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

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

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

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

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0so, with my equation,

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362709954712:dw

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362710284245:dw

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0oh the derivative gotcha

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362710497399:dw

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0should i just switch sides?

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

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

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

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

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0ya let me check real quick

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dangit, i need to stop rushing

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362711281571:dw

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362711568703:dw

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

zepdrix
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0yes, so I can simplify the left

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

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

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

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1362712446678:dw

zepdrix
 2 years 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
 2 years 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
 2 years 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.