UnkleRhaukus
  • UnkleRhaukus
\[{dy \over dx}={4y^2 \over x^2}+5{y \over x}+1\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
bookmark*
UnkleRhaukus
  • UnkleRhaukus
i think it is a Homogeneous Equation requiring a change of variables
UnkleRhaukus
  • UnkleRhaukus
(a linear shift

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