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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.

|dw:1362709077091:dw|

i suppose i should state what bernoulli is

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

|dw:1362709205198:dw|

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?

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

after you substitute everything.

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

ya, im just puzzled

so, with my equation,

|dw:1362709954712:dw|

ok that makes sense

|dw:1362710284245:dw|

oh the derivative gotcha

|dw:1362710497399:dw|

Yes good! :)

ok

should i just switch sides?

Just multiply everything by -1 :)

|dw:1362710598175:dw|

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

You had it in your middle step.

|dw:1362710722310:dw|

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

|dw:1362710969094:dw|

dangit, i need to stop rushing

haha nice

|dw:1362711281571:dw|

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

1/x

Cool c:

my bad

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

ok,

|dw:1362711568703:dw|

now, i want to isolate u' right?

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

yes, so I can simplify the left

ya sounds good c:

|dw:1362711903145:dw|

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.

|dw:1362712182784:dw|

Cx

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?

|dw:1362712446678:dw|

YESSS

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