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UnkleRhaukus
 5 years ago
\[{dy \over dx}={4y^2 \over x^2}+5{y \over x}+1\]
UnkleRhaukus
 5 years ago
\[{dy \over dx}={4y^2 \over x^2}+5{y \over x}+1\]

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UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0i think it is a Homogeneous Equation requiring a change of variables

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0I saw your work on it yesterday and saw no mistakes, so I'm curious as well

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5i believe we have to solve using y=mx where m= y/x

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0that's what rhaukus did yesterday, but he got the wrong answer... ideas imperialist?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trying a couple of things, I'll let you know if anything works.

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5m + x dm/dx = 4m^2 + 5 m +1

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5x dm /dx= 4m^2+4m+1

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5dx/x=dm/4m^2+4m+1

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5um then integrate :P

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5u can do from here?

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0ill give it a go ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Akshay is right, that's how you do it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is y = 1/2 x a solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a particular* solution.

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0\[ln(x)={1 \over 4x+2}+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\ln(x)=\frac{1}{4v+2}+C\] before you get too far without the negative sign and with x instead of v

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, \[y=\frac{1}{2}x\]is one solution. just let y = mx like akshay said, and when you plug that in you get an equation for m.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, m instead of v in what I said above, I always use v's for these types of questions!

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5lol i prefer m :P

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0i meant \[lnx=1/(4m+2) +c\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0\(y'/x=4u^2/x+5u/x+1\) where \(u=y/x\) \(y'/xu/x=4u^2/x+4u/x+1/x\) \(\frac{xy'y}{x^2}=4u^2/x+4u/x+1/x\) \(du/dx=4u^2/x+4u/x+1/x\) \(du/(4u^2+4u+1)=dx/x\) \(du/(2u+1)^2=dx/x\) \(\frac{1}{2}(2u+1)^{1}=\lnx+C\) \(\frac{1}{2}(\frac{2y}{x}+1)^{1}=\lnx+C\) Let see ...

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5just resubstitute the m now to get your equation.. i think we are done, aren't we?

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0\[{1 \over 2}(2{y \over x}+1)^{1}=lnx +C\] ... \[x+2(2y+x)lnkx=0\] where \[(e^C=k)\]

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5i think that your answer is correct :D

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0I thought you said it had an arctan in it Unk? what did you do wrong last time?

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5did u try wolfphram @turing?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going by unk's last post, no haven't looked at wolf looking at this http://openstudy.com/users/unklerhaukus#/updates/4f14e000e4b0b9109c92149b

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0\(1/(\lnx+C) =4(y/x)+2\) \(x/(\ln x+C)=4y+2x\) \(y=1/4( x/(\lnx+c) 2x)\)

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5@turing just check with wolfphram.. it is possible to have two different ways of represnting the same function

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0I just assumed they were too different, but I suppose you are right.

Akshay_Budhkar
 5 years ago
Best ResponseYou've already chosen the best response.5not a Guru at using wolhphram but i assume this is right http://www.wolframalpha.com/input/?i=dy%2Fdx%3D%284y%5E2%2Fx%5E2%29+%2B+5y%2Fx+%2B+1
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