/sci/ - Science & Math » Thread #10831961

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Anonymous Wed 24 Jul 04:07:15 2019 No.10831961 View Reply Original Report

Quoted By: >>10831968 >>10832002 >>10832004 >>10832007 >>10832031 >>10832034 >>10832058

Can /sci/ solve it? No it's not my homework this is more or less the /sci/ IQ test.

Anonymous

Anonymous Wed 24 Jul 2019 04:08:09 No.10831968 Report

Quoted By: >>10831978

>>10831961
I don’t feel like taking a derivative right now. How about some other time.

Anonymous

Anonymous Wed 24 Jul 2019 04:12:03 No.10831978 Report

Quoted By:

>>10831968
Suit yourself, I'm experimenting with this thread. Might start posting more equations of different types.

Anonymous

Anonymous Wed 24 Jul 2019 04:18:54 No.10831997 Report

Quoted By:

let y = vx

Anonymous

Anonymous Wed 24 Jul 2019 04:20:55 No.10832002 Report

Quoted By:

>>10831961
it is solved? what are you even asking for?

Anonymous

Anonymous Wed 24 Jul 2019 04:21:47 No.10832004 Report

Quoted By:

>>10831961
this looks like radial coordinates or some kind of change of variable is needed. im done by my standards

Anonymous

Anonymous Wed 24 Jul 2019 04:23:51 No.10832007 Report

Quoted By: >>10832547

>>10831961
>can /sci/ solve this?
>doesn't even say what he's asking for
brainlet.

Anonymous

Anonymous Wed 24 Jul 2019 04:37:17 No.10832027 Report

Quoted By: >>10832132

>dy/dx
I see you, you sneaky bastard.

Anonymous

Anonymous Wed 24 Jul 2019 04:39:00 No.10832031 Report

Quoted By:

>>10831961
The family of functions implicitly defined by the curve $\frac{x^3}{3}-xy^2 = c$ ?

Anonymous

Anonymous Wed 24 Jul 2019 04:42:08 No.10832034 Report

Quoted By:

>>10831961

x^6 - 2x^2 +1

Anonymous

Anonymous Wed 24 Jul 2019 04:58:29 No.10832058 Report

Quoted By: >>10832093

>>10831961
Let u = x y^2

du/dx = y^2 + 2 x y dy/dx
= y^2 + (x^2 - y^2)
= x^2

u = (1/3) x^3 + C)
y = sqrt(u/x) = sqrt((1/3) x^2 + C/x)

Anonymous

Anonymous Wed 24 Jul 2019 05:14:44 No.10832093 Report

Quoted By:

>>10832058
or you can solve it as a Bernoulli equation

$\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{1}{2x}y = \frac{x}{2y} [\math]<br /><br />multiply by <script type="math/tex">2y[\math]<br /><br /><script type="math/tex">2y\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{1}{2x}2y^2 = x[\math]<br /><br />substitute <script type="math/tex">w = y^2[\math] (therefore <script type="math/tex">\frac{\mathrm{d}w}{\mathrm{d}x} = 2y\frac{\mathrm{d}y}{\mathrm{d}x}[\math])<br /><br /><script type="math/tex">\frac{\mathrm{d}w}{\mathrm{d}x} + \frac{1}{x}w = x[\math]<br /><br />and solve as a simple 1st order equation$

Anonymous

Anonymous Wed 24 Jul 2019 05:43:01 No.10832132 Report

Quoted By:

>>10832027
Sneaky how?

Anonymous

Anonymous Wed 24 Jul 2019 10:12:07 No.10832547 Report

Quoted By:

>>10832007
What do you mean asking for kek it's pretty straight forward if you've done baby calc

Anonymous

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Anonymous Wed 24 Jul 2019 21:54:33 No.10833708 Report

Quoted By:

>another IQ thread

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