Question

LOKI!!!!!! HELPPPPPPP :P

Answer 1

LOL!

Answer 2

shut up lololol

Answer 3

alright =P ^=^

Answer 4

he isint helping -/-

Answer 5

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

Answer 6

lmao

Answer 7

that's loki :) lol

Answer 8

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

Answer 9

it would really help

Answer 10

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

Answer 11

lol, loki is here =D

Answer 12

Where's your initial condition?

Answer 13

LOKI

Answer 14

Hello

Answer 15

its 0,1

Answer 16

y(0)=1

Answer 17

okay, give me a few minutes to sort out something

Answer 18

What don't you get?

Answer 19

lemmy upload a pic

Answer 20

you actualy took REAL pics, wow

Answer 21

and my other option was... fake ones?

Answer 22

lol

Answer 23

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...

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

i believe the seen point in 0,1

Answer 28

then again i have no idea what a seed point is

Answer 29

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

Answer 30

Seed point here is y(0)=1

Answer 31

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

Answer 32

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.

Answer 33

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

Answer 34

where did that whole formula come from? O.o

Answer 35

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

Answer 36

From this ^^

Answer 37

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 -.-

Answer 38

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.

Answer 39

that was a big waste of time -.-

Answer 40

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

Answer 41

You panicked.

Answer 42

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

Answer 43

np sstarica :)

Answer 44

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

Answer 45

night

Answer 46

isisnt it like midnight for u guys?

Answer 47

1:35 am

Answer 48

So you right with this one now?

Answer 49

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

Answer 50

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