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

- UnkleRhaukus

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

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- TuringTest

bookmark*

- UnkleRhaukus

i think it is a Homogeneous Equation requiring a change of variables

- UnkleRhaukus

(a linear shift

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## More answers

- TuringTest

I saw your work on it yesterday and saw no mistakes, so I'm curious as well

- Akshay_Budhkar

i believe we have to solve using
y=mx
where m= y/x

- TuringTest

that's what rhaukus did yesterday, but he got the wrong answer...
ideas imperialist?

- Akshay_Budhkar

dy=mdx + xdm

- anonymous

I'm trying a couple of things, I'll let you know if anything works.

- UnkleRhaukus

thanks

- Akshay_Budhkar

m + x dm/dx = 4m^2 + 5 m +1

- Akshay_Budhkar

x dm /dx= 4m^2+4m+1

- Akshay_Budhkar

dx/x=dm/4m^2+4m+1

- Akshay_Budhkar

um then integrate :P

- Akshay_Budhkar

u can do from here?

- UnkleRhaukus

ill give it a go ,

- anonymous

Akshay is right, that's how you do it

- anonymous

is y = -1/2 x a solution?

- anonymous

a particular* solution.

- UnkleRhaukus

\[ln(x)={1 \over 4x+2}+c\]

- anonymous

\[\ln(x)=\frac{-1}{4v+2}+C\] before you get too far without the negative sign and with x instead of v

- anonymous

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.

- anonymous

Sorry, m instead of v in what I said above, I always use v's for these types of questions!

- Akshay_Budhkar

lol i prefer m :P

- UnkleRhaukus

i meant \[lnx=-1/(4m+2) +c\]

- anonymous

Me too! I prefer 'V'

- watchmath

\(y'/x=4u^2/x+5u/x+1\) 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 ...

- Akshay_Budhkar

just re-substitute the m now to get your equation.. i think we are done, aren't we?

- UnkleRhaukus

\[{-1 \over 2}(2{y \over x}+1)^{-1}=ln|x| +C\]
...
\[x+2(2y+x)ln|kx|=0\]
where \[(e^C=k)\]

- Akshay_Budhkar

i think that your answer is correct :D

- TuringTest

I thought you said it had an arctan in it Unk?
what did you do wrong last time?

- Akshay_Budhkar

did u try wolfphram @turing?

- TuringTest

I'm going by unk's last post, no haven't looked at wolf
looking at this
http://openstudy.com/users/unklerhaukus#/updates/4f14e000e4b0b9109c92149b

- watchmath

\(-1/(\ln|x|+C) =4(y/x)+2\)
\(-x/(\ln |x|+C)=4y+2x\)
\(y=1/4( -x/(\ln|x|+c) -2x)\)

- Akshay_Budhkar

@turing just check with wolfphram.. it is possible to have two different ways of represnting the same function

- TuringTest

I just assumed they were too different, but I suppose you are right.

- Akshay_Budhkar

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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