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

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

Mathematics
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i think it is a Homogeneous Equation requiring a change of variables
(a linear shift

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I saw your work on it yesterday and saw no mistakes, so I'm curious as well
i believe we have to solve using y=mx where m= y/x
that's what rhaukus did yesterday, but he got the wrong answer... ideas imperialist?
dy=mdx + xdm
I'm trying a couple of things, I'll let you know if anything works.
thanks
m + x dm/dx = 4m^2 + 5 m +1
x dm /dx= 4m^2+4m+1
dx/x=dm/4m^2+4m+1
um then integrate :P
u can do from here?
ill give it a go ,
Akshay is right, that's how you do it
is y = -1/2 x a solution?
a particular* solution.
\[ln(x)={1 \over 4x+2}+c\]
\[\ln(x)=\frac{-1}{4v+2}+C\] before you get too far without the negative sign and with x instead of v
yeah, \[y=-\frac{1}{2}x\]is one solution. just let y = mx like akshay said, and when you plug that in you get an equation for m.
Sorry, m instead of v in what I said above, I always use v's for these types of questions!
lol i prefer m :P
i meant \[lnx=-1/(4m+2) +c\]
Me too! I prefer 'V'
\(y'/x=4u^2/x+5u/x+1\) where \(u=y/x\) \(y'/x-u/x=4u^2/x+4u/x+1/x\) \(\frac{xy'-y}{x^2}=4u^2/x+4u/x+1/x\) \(du/dx=4u^2/x+4u/x+1/x\) \(du/(4u^2+4u+1)=dx/x\) \(du/(2u+1)^2=dx/x\) \(-\frac{1}{2}(2u+1)^{-1}=\ln|x|+C\) \(-\frac{1}{2}(\frac{2y}{x}+1)^{-1}=\ln|x|+C\) Let see ...
just re-substitute the m now to get your equation.. i think we are done, aren't we?
\[{-1 \over 2}(2{y \over x}+1)^{-1}=ln|x| +C\] ... \[x+2(2y+x)ln|kx|=0\] where \[(e^C=k)\]
i think that your answer is correct :D
I thought you said it had an arctan in it Unk? what did you do wrong last time?
did u try wolfphram @turing?
I'm going by unk's last post, no haven't looked at wolf looking at this http://openstudy.com/users/unklerhaukus#/updates/4f14e000e4b0b9109c92149b
\(-1/(\ln|x|+C) =4(y/x)+2\) \(-x/(\ln |x|+C)=4y+2x\) \(y=1/4( -x/(\ln|x|+c) -2x)\)
@turing just check with wolfphram.. it is possible to have two different ways of represnting the same function
I just assumed they were too different, but I suppose you are right.
not 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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