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

- anonymous

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

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- schrodinger

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- anonymous

Yes, this is a bit strange because they want me to substitute u=y^(1-n)

- anonymous

so, im thinking to start off, i will get x and y on one side.

- anonymous

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## More answers

- anonymous

i suppose i should state what bernoulli is

- 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

Woops, \(\large u'=-y^{-2}y'\)

- anonymous

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- abb0t

Haven't seen bernoulli in some time.

- 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

u' = y^-2y' is what your going to substitute. essentially, you're going to get a linear differential euqation.

- abb0t

You should get something like:
u' + p(x)u=g(x)

- abb0t

after you substitute everything.

- 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

I tried to use colors to make that clearer, but it's still rather confusing lol

- anonymous

ya, im just puzzled

- anonymous

so, with my equation,

- 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

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- 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

ok that makes sense

- anonymous

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- 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

oh the derivative gotcha

- 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

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- zepdrix

Yes good! :)

- zepdrix

Remember that before you can get an integrating factor and all that fun stuff, your leading term needs to be positive.

- anonymous

ok

- anonymous

should i just switch sides?

- 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

Just multiply everything by -1 :)

- anonymous

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- zepdrix

Woops, standard form looks like this,\[\large u'+P(x)u=Q(x)\]

- zepdrix

You had it in your middle step.

- anonymous

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- zepdrix

heh XD

- anonymous

ok so, i integrate both sides?

- zepdrix

No you need to find an integrating factor \(\large \mu=e^{\int P(x)\;dx}\)
Look familiar I hope? :)

- anonymous

ya let me check real quick

- anonymous

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- 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

dangit, i need to stop rushing

- 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

Yes you do need to stop rushing Chris Farley!! Slow down in life! Or you'll end up leaving us too soon......... :'C

- anonymous

haha nice

- anonymous

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- zepdrix

\[\huge e^{-\ln x} \qquad = \qquad e^{\ln (x^{-1})} \qquad = \qquad ?\]

- anonymous

1/x

- zepdrix

Cool c:

- anonymous

my bad

- anonymous

ok so now that i have that, i multiply by x^-1

- 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

ok,

- anonymous

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- anonymous

now, i want to isolate u' right?

- 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

From here, you want to recognize that the LEFT side is the result of the product rule.

- 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

yes, so I can simplify the left

- zepdrix

ya sounds good c:

- anonymous

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- zepdrix

Oh sorry I forget you were calling it \(\large I\). I hope I didn't make that more confusing c:

- zepdrix

yah everything looks good so far.

- anonymous

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- anonymous

Cx

- zepdrix

Yes good. Then we have to do a little work changing our \(\large u\) back to \(\large y\)

- anonymous

ok, u=y^-1

- zepdrix

Good.\[\large \frac{1}{y}=2x \ln x+Cx\]So how do we solve for y?

- anonymous

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- 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

YESSS

- anonymous

thank you so much for your patience, i have been doing very well in my class because of your help

- 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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