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/sci/ - Science & Math

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>> No.4801292 [View]

>>4801289
Thanks. Oh textbook, how could you lie to me.

>> No.4801281 [View]

>>4801274

Interestingly,

<span class="math"> \frac{2x}{Cx +1}[/spoiler]

>> No.4801270 [View]

>>4801258
Even if that were the case, Wolfram says otherwise.

>> No.4801253 [View]

>>4801246

Ah? Care to explain?

>> No.4801249 [View]

>>4801241
Alright, will do.

>> No.4801232 [View]
File: 178 KB, 611x1200, little_regent_by_GunnerRomantic.jpg [View same] [iqdb] [saucenao] [google]
4801232

Quick question on first-order ordinary differential equations:

<span class="math">2x^2 y' = y^2[/spoiler]

<span class="math">(=) y' = \frac {1}{2x^2} * y^2 [/spoiler]

<span class="math">(=) \frac{dy}{dx} = \frac {1}{2x^2} * y^2[/spoiler]

<span class="math">(=) \frac {1}{y^2} dy = \frac{1}{2x^2} dx[/spoiler]

<span class="math">(=) \int \frac{1}{y^2} dy = \int \frac{1}{2x^2} dx[/spoiler]

[...]

Then I'd just calculate the integrals and solve for y. I get the result

<span class="math">\frac {2x}{1 -2xC}[/spoiler],

which is not correct, though. Is the mistake already somewhere in the steps I posted above, and if yes, what is it?

Thanks!

>> No.4792849 [View]

>>4792813
What do you mean by "interested" and "uninterested"? Interested in what (or whom)?

>> No.4792779 [View]
File: 458 KB, 842x1263, johnnie_walker_wisdom_by_gunnerromantic-d397qpm.jpg [View same] [iqdb] [saucenao] [google]
4792779

Do people of similiar intelligence levels stick together, given that there's a proper definition of "intelligence" in the first place, and if yes, why is that?

>> No.4786640 [View]

>>4786628
That's the kind of hint I needed. Many thanks!

>> No.4786622 [View]

>>4786615
The syntax is very similiar to the usual LaTeX syntax, and is used by typing math and /math in square brackets. It's a kind of programming language for mathematical texts etc.

>> No.4786608 [View]
File: 34 KB, 500x390, christiane-f-wir-kinder-vom-bahnhof-zoo-143018l.jpg [View same] [iqdb] [saucenao] [google]
4786608

Hi,
I hope someone can help me with a (probably rather easy) integration problem. I just don't see the trick that's needed, I suppose, plus the task requires me to use partial integration as opposed to substitution or any other method, although other methods might be suited better.

<span class="math"> \int \sin^3(x) * \cos(x) dx[/spoiler]

My attempt was to set

a' = <span class="math"> \cos(x) [/spoiler]
a = <span class="math"> \sin(x) [/spoiler]
b = <span class="math"> \sin^3(x) [/spoiler]
b' = <span class="math"> 3\sin^2(x) * \cos(x) [/spoiler]

so

<span class="math"> \int \sin^3(x) * \cos(x) dx = \sin^4(x) - \int \sin(x) * (3(\sin^2(x)) * \cos(x)) dx [/spoiler]

Sadly, that doesn't help me much. Even if I partially integrate further, it just becomes a terrible mess. Any hints? I don't need a complete solution, just a hint would be helpful.

>> No.4784831 [View]

Thanks for the useful posts!

>> No.4784576 [View]

>>4784447
It seems like you didn't read my third post - wasn't me.

Normally I'd use a computer, but in our exams (it's fairly early in the curriculum), we aren't allowed to because they want us to know the basics first.

>> No.4784442 [View]

>>4784440
Yeah, that's what thought. I'll just learn all of the trigonometric derivations first, I suppose..

>> No.4784437 [View]

Bumping.

>>4784430
Also, I'll just be using a tripcode because some people think that trolling on 4chan has ever been funny.

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