## anonymous 5 years ago LOKI!!!!!! HELPPPPPPP :P

1. anonymous

LOL!

2. anonymous

shut up lololol

3. anonymous

alright =P ^=^

4. anonymous

he isint helping -/-

5. anonymous

he's prolly drinking, eating , helping the gf or day dreaming lol

6. anonymous

lmao

7. anonymous

that's loki :) lol

8. anonymous

I wish if there was a nudge button here to nudge the person you want to talk to :(

9. anonymous

it would really help

10. anonymous

i dont think anyone else here can do differential equations :(

11. anonymous

lol, loki is here =D

12. anonymous

Where's your initial condition?

13. anonymous

LOKI

14. anonymous

Hello

15. anonymous

its 0,1

16. anonymous

y(0)=1

17. anonymous

okay, give me a few minutes to sort out something

18. anonymous

What don't you get?

19. anonymous

lemmy upload a pic

20. anonymous

21. anonymous

you actualy took REAL pics, wow

22. anonymous

and my other option was... fake ones?

23. anonymous

lol

24. anonymous

well... it is just that you are the first person I see taking photos of your exercises and uploading them here, but no, you had no other options...

25. anonymous

You don't think they just want you to solve the equation exactly for comparison? Family of solutions is all solutions with the constant intact.

26. anonymous

Because for Euler's Method to work, you need a seed point.

27. anonymous

i cant techniclly solve the equation, i do know other techniques of diff equation solving but i am supposed to, at the current point in time, know nothing more than eulers approximation and separation of variables

28. anonymous

i believe the seen point in 0,1

29. anonymous

then again i have no idea what a seed point is

30. anonymous

yeah, you can use sep. vars. to solve this. You end up with $x^2-y^2=c$which is a family of hyperbola.

31. anonymous

Seed point here is y(0)=1

32. anonymous

hows you get x2-y2-c?????

33. anonymous

If you look at that curve, you'll see that, when x=0, you have$0-(y(0))^2=c \rightarrow y(0)=\pm \sqrt{-c}$as a different seed.

34. anonymous

The c would be less than or equal to zero (real solutions here).

35. anonymous

where did that whole formula come from? O.o

36. anonymous

$\frac{dy}{dx}=\frac{x}{y}\rightarrow y dy = x dx \rightarrow \frac{y^2}{2}=\frac{x^2}{2}+c \rightarrow x^2-y^2=c$

37. anonymous

From this ^^

38. anonymous

oh wow... separation of variables does work on this one... ive been confusing this with the last problem which was y'=x-y where seoaration doesnt work -.-

39. anonymous

Basically, then, since the family of curves is defined by c, you can pick your y(0) to generate a new hyperbola within the family.

40. anonymous

that was a big waste of time -.-

41. anonymous

Your point would be (0,A) where A is 'Arman's choice'.

42. anonymous

You panicked.

43. anonymous

I myself have actually learned something ._. thank you loki

44. anonymous

np sstarica :)

45. anonymous

^_^ alright, off to bed soon, good night :)

46. anonymous

night

47. anonymous

isisnt it like midnight for u guys?

48. anonymous

1:35 am

49. anonymous

So you right with this one now?

50. anonymous

k arman, i gotta go...my eyes are burning. see you round ;)

51. anonymous

my bad for not replying, i was grpahing it ><