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

Question

turingtest · Answer

bookmark*

unklerhaukus · Answer

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

unklerhaukus · Answer

(a linear shift

turingtest · Answer

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

akshay_budhkar · Answer

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

turingtest · Answer

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

akshay_budhkar · Answer

dy=mdx + xdm

anonymous · Answer

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

unklerhaukus · Answer

thanks

akshay_budhkar · Answer

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

akshay_budhkar · Answer

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

akshay_budhkar · Answer

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

akshay_budhkar · Answer

um then integrate :P

akshay_budhkar · Answer

u can do from here?

unklerhaukus · Answer

ill give it a go ,

anonymous · Answer

Akshay is right, that's how you do it

anonymous · Answer

is y = -1/2 x a solution?

anonymous · Answer

a particular* solution.

unklerhaukus · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

akshay_budhkar · Answer

lol i prefer m :P

unklerhaukus · Answer

i meant $$lnx=-1/(4m+2) +c$$

anonymous · Answer

Me too! I prefer 'V'

watchmath · Answer

$y'/x=4u^2/x+5u/x+1$ 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 · Answer

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

unklerhaukus · Answer

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

akshay_budhkar · Answer

i think that your answer is correct :D

turingtest · Answer

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

akshay_budhkar · Answer

did u try wolfphram @turing?

turingtest · Answer

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 · Answer

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

akshay_budhkar · Answer

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

turingtest · Answer

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

akshay_budhkar · Answer

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