LOKI!!!!!! HELPPPPPPP :P

- anonymous

LOKI!!!!!! HELPPPPPPP :P

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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- anonymous

LOL!

- anonymous

shut up lololol

- anonymous

alright =P ^=^

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## More answers

- anonymous

he isint helping -/-

- anonymous

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

- anonymous

lmao

- anonymous

that's loki :) lol

- anonymous

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

- anonymous

it would really help

- anonymous

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

- anonymous

lol, loki is here =D

- anonymous

Where's your initial condition?

- anonymous

LOKI

- anonymous

Hello

- anonymous

its 0,1

- anonymous

y(0)=1

- anonymous

okay, give me a few minutes to sort out something

- anonymous

What don't you get?

- anonymous

lemmy upload a pic

- anonymous

- anonymous

you actualy took REAL pics, wow

- anonymous

and my other option was... fake ones?

- anonymous

lol

- 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...

- 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.

- anonymous

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

- 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

- anonymous

i believe the seen point in 0,1

- anonymous

then again i have no idea what a seed point is

- 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.

- anonymous

Seed point here is y(0)=1

- anonymous

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

- 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.

- anonymous

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

- anonymous

where did that whole formula come from? O.o

- anonymous

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

- anonymous

From this ^^

- 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 -.-

- 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.

- anonymous

that was a big waste of time -.-

- anonymous

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

- anonymous

You panicked.

- anonymous

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

- anonymous

np sstarica :)

- anonymous

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

- anonymous

night

- anonymous

isisnt it like midnight for u guys?

- anonymous

1:35 am

- anonymous

So you right with this one now?

- anonymous

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

- anonymous

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

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