Question

Okay I know this is a homogenous equation and supposed to use u substitution (u= x/y) but it's just too convoluted.
y' = y/x + x / [(x^2 * y)^1/3 + (xy)^1/2 + x ]
Thank you in advance for any help.

Answer 1

i get\[x=\frac{1}{u'(u^{\frac{1}{3}}+u^{\frac{1}{2}}+1)}\]

Answer 2

okay I see you now :D thanks a lot, but how did you just sub u in like that

Answer 3

for the first part of the denom its x^2y and then xy so they both can't equal u can they?

Answer 4

or you can write it as\[u'(u^{\frac{1}{3}}+u^{\frac{1}{2}}+1)=\frac{1}{x}\]

Answer 5

Once its in that form I can take it form there, just use distributive property on u' and then simply integrate, but I just don't know how to get it there.

Answer 6

\[(x^2y)^{\frac{1}{3}}\] right?

Answer 7

yeah

Answer 8

so if u = y/x => x^3u = x^2y

Answer 9

do you mean (x^3)u or x^(3u)
I dunno why I'm having such a hard time wrapping my head around this lol

Answer 10

ohhhh derp got it (x^3)u obviously

Answer 11

So then I can take the cubed root and take x out of the picture, and u stays cube rooted. I'm starting to get somewhere :D

Answer 12

yes
and xy = x^2 u

Answer 13

Okay well that was the part I was stuck on, thank you so much man.. medal well earned you should be able to get double in the tougher maths :P

Answer 14

lol! glad i could help!

Answer 15

Then just a simple separable equation :D 3 more days till S5:E2 of archer!!

Answer 16

that's right! there are a couple of sites which give some tidbits about the season but the closing montage of E1 gave a preview as well.
can't wait!!!

Answer 17

Haven't seen anything besides the closing montage but that was exciting enough :D I'm gonna get off the site for a bit though I have work at 4, great meeting you :) Really cool that you come here and help people learn. (and not just give them answers like me >_> haha) Take care!

Answer 18

you too!!! have fun at work!