Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

xy' + y = −2xy^2 Bernoulli Differential equation.

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

Yes, this is a bit strange because they want me to substitute u=y^(1-n)
so, im thinking to start off, i will get x and y on one side.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i suppose i should state what bernoulli is
\[\large xy' + y = −2xy^2\]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 \(\large u'=-y^{-2}\)
Woops, \(\large u'=-y^{-2}y'\)
Haven't seen bernoulli in some time.
ya im a bit shaky on this but im caught up to you zep, so the u'=y'y^-2 , can you explain that?
u' = y^-2y' is what your going to substitute. essentially, you're going to get a linear differential euqation.
You should get something like: u' + p(x)u=g(x)
after you substitute everything.
\[\large y'+p(x)y=q(x)y^n\] 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 :)
I tried to use colors to make that clearer, but it's still rather confusing lol
ya, im just puzzled
so, with my equation,
\[\large xy' + y = −2xy^2\]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 \(\large u\).
Our \(\large n\) in this case is 2, because our y^n was the term on the RIGHT side of the equation. \(\large u=y^{1-n} \qquad \rightarrow \qquad u=y^{2-1}\qquad \rightarrow \qquad u=y^{-1}\)
ok that makes sense
I think maybe you missed a negative when taking the derivative of your u. \(\huge u=y^{-1} \qquad \rightarrow \qquad u'=\color{red}{-}y^{-2}y'\)
oh the derivative gotcha
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
Yes good! :)
Remember that before you can get an integrating factor and all that fun stuff, your leading term needs to be positive.
should i just switch sides?
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.
Just multiply everything by -1 :)
Woops, standard form looks like this,\[\large u'+P(x)u=Q(x)\]
You had it in your middle step.
heh XD
ok so, i integrate both sides?
No you need to find an integrating factor \(\large \mu=e^{\int P(x)\;dx}\) Look familiar I hope? :)
ya let me check real quick
Woops the negative sign is also part of your \(\large P(x)\). Recall that the standard form shows addition in the middle there.
dangit, i need to stop rushing
And your equation should be \(\large u'-\dfrac{1}{x}u=2\) Positive 2 on the right. I think we missed that earlier.
Yes you do need to stop rushing Chris Farley!! Slow down in life! Or you'll end up leaving us too soon......... :'C
haha nice
\[\huge e^{-\ln x} \qquad = \qquad e^{\ln (x^{-1})} \qquad = \qquad ?\]
Cool c:
my bad
ok so now that i have that, i multiply by x^-1
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.
now, i want to isolate u' right?
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|
From here, you want to recognize that the LEFT side is the result of the product rule.
\[\large \left[u(x)f(x)\right]' \qquad = \qquad u'(x)f(x)+u(x)f'(x)\] We do that have in this case don't we? Where our f(x)=x^{-1}
yes, so I can simplify the left
ya sounds good c:
Oh sorry I forget you were calling it \(\large I\). I hope I didn't make that more confusing c:
yah everything looks good so far.
Yes good. Then we have to do a little work changing our \(\large u\) back to \(\large y\)
ok, u=y^-1
Good.\[\large \frac{1}{y}=2x \ln x+Cx\]So how do we solve for y?
Yay good job! I threw it into Wolfram just a moment ago to make sure we didn't make any fatal errors. Same result! Yay! :)
thank you so much for your patience, i have been doing very well in my class because of your help
Oh that's good to hear! c: I'm sure it's not just me. You seem to understand these concepts very very quickly.

Not the answer you are looking for?

Search for more explanations.

Ask your own question