/sci/ - Science & Math

[Anonymous No.2505278]

Hey, can someone explain integration via separations of variables to me?

pic unrelated

[Anonymous No.2505286]

First, you must create the universe.

[Anonymous No.2505297]

Then you stick your dick in some apple pie.

[Anonymous No.2505302]

Integrating by separation of variables? Are you sure you mean integrating, and not solving PDEs?

[Anonymous No.2505304]

y' - x/2 = 3
y' = 3+x/2
y = 3x+x^2 + C

[Anonymous No.2505310]

Finally you crack open a book and brute force some problems so you get a decent list in your mind of which variables to separate out of your problem.

There aren't many at the introductory level.

[Anonymous No.2505314]

>>2505304
Integrated wrong, but still the same concept

[Anonymous No.2505317]

>>2505302
I believe I mean integrating.

ex: dy/dx = xy

[Anonymous No.2505329]

The trick is to act like dy and dx or any variants can act separately, even though they're just notation.

[Anonymous No.2505341]

>>2505278
I'd like to find the area under her curves.

>giggity...

[Anonymous No.2505343]

>>2505317

dy/dx = xy

1/y dy/dx = x

1/y dy = x dx (Integrate)

lny = (x^2)/2 + C

y = e^((x^2)/2 + C)

Overall: Treat dy/dx as a fraction (It's not)

[Anonymous No.2505349]

if yoy have something like dy\dx = f(x,y) or dy = f(x,y)dx
you try to write it in the form
g(y)dy = h(x)dx whith g and h ONLY dependent on y and x respectively, then integrate both sides
<div class="math"> int g(y) dy = int h(x) dx </div>
and you have the answer after getting y alone

[Anonymous No.2505361]

<div class="math"> \int g(y) dy = \int h(x) dx</div>

[Anonymous No.2505387]

OP here: I think I understand it better, but this problem is frustrating me.

I can't figure out how to get all the Y terms together.

[Anonymous No.2505421]

bump?

[Anonymous No.2505422]

>>2505387
<div class="math"> \frac{dy}{\sqrt{y} }= \frac{4 \ln(x)}{x} dx</div>

[Anonymous No.2505428]

dy/sqrt(y)=4ln(x)/x dx

[Anonymous No.2505447]

your problem is algebra

not integration

if you can't separate the x from the y there

[Anonymous No.2505474]

>>2505447
Not really.

After I separate the two and attempt to integrate is where I get lost.
dy/sqrt(y)= 4ln(x)/x
2sqrt(y) = 2ln(x)^2
y = ln(x)^4
1 = ln(e)^4 + C
1=C?

[Anonymous No.2505483]

>>2505474 here:
I obviously did that wrong, but I don't know where it went awry.

[Anonymous No.2505498]

y = ln(x)^4 + c

and c = 1

so
y = ln(x)^4 + 1

[Anonymous No.2505508]

>>2505498
according to my book that is wrong.

Anyone have any ideas?

[Anonymous No.2505533]

RAGE, come on...

[Anonymous No.2505946]

>>2505533
>>2505498
ITT: people who think ln(e^(1))=0

The answer is just ln(x)^4