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

1. TuringTest

bookmark*

2. UnkleRhaukus

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

3. UnkleRhaukus

(a linear shift

4. TuringTest

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

5. Akshay_Budhkar

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

6. TuringTest

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

7. Akshay_Budhkar

dy=mdx + xdm

8. anonymous

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

9. UnkleRhaukus

thanks

10. Akshay_Budhkar

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

11. Akshay_Budhkar

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

12. Akshay_Budhkar

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

13. Akshay_Budhkar

um then integrate :P

14. Akshay_Budhkar

u can do from here?

15. UnkleRhaukus

ill give it a go ,

16. anonymous

Akshay is right, that's how you do it

17. anonymous

is y = -1/2 x a solution?

18. anonymous

a particular* solution.

19. UnkleRhaukus

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

20. anonymous

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

21. 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.

22. anonymous

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

23. Akshay_Budhkar

lol i prefer m :P

24. UnkleRhaukus

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

25. anonymous

Me too! I prefer 'V'

26. 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 ...

27. Akshay_Budhkar

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

28. UnkleRhaukus

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

29. Akshay_Budhkar

30. TuringTest

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

31. Akshay_Budhkar

did u try wolfphram @turing?

32. 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

33. watchmath

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

34. Akshay_Budhkar

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

35. TuringTest

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

36. 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