if a differential equations solution divides by x^2 and the initial condition is a point where x=0, what do you do? (problem: dy/dx=xyy or dy/dx=xy^2)

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if a differential equations solution divides by x^2 and the initial condition is a point where x=0, what do you do? (problem: dy/dx=xyy or dy/dx=xy^2)

Mathematics
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initial condition is (0,-2)
You can use separation by variables here and then integrate. Once you have y after integration, you can use the initial condition.
i did that, but my solution is y=(-2/x^2)+2 which doesnt technically do through the initial condition

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Other answers:

i think its a vertical asymptote which... definitely isint the point XD
There's nothing wrong with that equation. After you put it into separable form and solve, you should get,\[\frac{x^2y}{2}+cy+1=0\]Initial condition,\[0-2c+1=0 \rightarrow c=\frac{1}{2}\]
You can then factor out y and rearrange into the form y=y(x).
brb
i got c=2?
When you separated the variable and tried to get y by itself did you include c and also multiply the y by it?
.... -.-
ima try that, how about: y'=1+yy
You can do that problem the same way. I look at it as y' is dy/dx and then separate variables. Make sure when you do though, that the 1 goes with the y^2. So move the entire quantity.
dang, the small things get by me... that should be something like arctany=x+c? so tan(x+c)=y?
that explains why my condition is in radians
Yes.
alright... major confusion... \[y'=(e ^{x-y})\div(1+e ^{x})\]
You want to solve that?
sadly ><
Just pull out the e^(-y) and it becomes separable ;)
asoghaqoeraifoiafauirfhlanihusdfvnnfldanufd
\[e^ydy=\frac{e^x}{1+e^x}dx\]
differential equations are gay -.-
lol
dont tell me any more!!!
tell you what? answers?
yeah.. dont tell me any more lol
no worries :D
I think you have the best icon...if that helps...
hahahaha it was annoyingly hard to make XD
i have a pretty good feeling that your icon is not you... though there is a significant chance that i am wrong given by the circumstances
it could be me, since i look like that
lol, how did your frigging doughnut go?
pretty good, i have a way of doing things when i need help: use the first sentence or line of the help, fail a thousand times before i use the second, so on and so forth. so i used quite a bit, but not all of your help in the process of doing the assignment, and in the end i was flowing through it quite nicely, i compared my results to yours and they were identical, i then compared my results to the assignment and apparently the last couple steps, which i had done by hand, i could take the integral with a calculator... and then later he gave us the hand written answer, so i felt proud that i did it without a calculator... but yeah your help is insanely apreciated
alright im stuck at e^y=ln(1+e^x)+c initial condition is (0,1)
Is that y supposed to be y' (y-prime)?
nope, thats my solution
to the above equation
oh..
Yeah. You're right. Leave it like that too.
but i need to find c
Oh, just plug your numbers in.
\[e=\ln 2 +c \rightarrow c = e- \ln 2\]
any way to do that without a calc?
You mean simplify?
No.
What's wrong with Euler's number? :p
ah god its 1... can you check these real quick?
Check what?
Do you ever sleep?
1 Attachment
not too much anymore lol
I need paper...I've run out...hang on...
but all my work is shown?
I get something different for the first one. Just go to bed and I'll look at it and scan and you can look at it when you get up.
do i have to multiply the c by 2 aswell?
I'll do it here:
my answer doesnt make sense anyways... when i find y' its all weird and quotienty
\[\frac{dy}{dx}=xy^2 \rightarrow y^{-2}dy = x dx \rightarrow -y^{-1}=x+c\]
Your boundary condition is (0,-2), so
x+c or (1/2)xx+c?
\[-2^{-1}=0+c \rightarrow c=-\frac{1}{2}\]
So\[y^{-1}+x=\frac{1}{2} \rightarrow 2+2xy=y \rightarrow y(1-2x)=2 \rightarrow y=\frac{2}{1-2x}\]
waitttt the integral of xdx is (1/2)xx+c though!
Oh crap, I forgot the frigging x. Sorry Arman.
nah dont be sorry, i should hit the sheets tho, i'll probably have more questions tomorrow lol
I still get a different answer to you.
basic differential equations (like these) arent usually part of calc II are they?
Yours is close. I'll work through them - they won't take long - and scan and you can read it in the morning.
different answer?
I'm not sure about the American system.
y=2/(1-x^2)
did you use (0,-2) for the initial whatever?
pellet, I wrote it wrong on my paper. I wrote (0,2). SORRY AGAIN :'(
NOW I get what you got.
>< nah man i hate when ppl tell me sorry
lol XD
my calc teacher, this morning, was like "oh flutter..." *covers mouth*... *whole class starts laughing caus we hear him say pellet sometimes but never the f word so it meant he was really screwed over XD
lol, it's great when that happens. I got the same results as you for the other two. The *only* thing I would do differently is leave \[e-\log 2\]as is, and not use a numerical approximation (in the last one).
thats probably better in the math world, where you need to be precise
Yeah, exactly.
is that prefered in college?
preferred*
Absolutely. You'd only use a numerical approximation for the time you actually need to extract a number. Other than that, we leave it because we may want to do further mathematics on it and we lose information/exactness if we start approximating beforehand.
just like the AP exams lol. i think i asked you before but what are the remaining major math courses after differential eqas and linear algebra (when i say top i mean like... there should be less than 3 "top" ones)
?
Yeah, I really don't know. Your system is different to the system in UK, Australia and Canada. We don't break mathematics up into sections like calculus and then apply a skill level to it. We apply a skill level to ALL of mathematics at particular points and you learn EVERYTHING at a particular level.
oh... well what are the chances that i can attain a college grad level (in math) of mathematical understanding in the next year?
given the rate that i covered calc 1 and 2 within 1.5 months
I'd say you're pretty good.
College grad level in calculus...is that what you're asking?
You have to cover calculus of a complex variable as well. There's also tensor calculus.
not just calculus, but all mathematics, can i attain that sort of knowledge, about 4 or more years of college math in the next 9 months?
complex variable wont be too hard, what is tensor calc??
It *might* take a little longer ;p
They're devices we use to extend scalars, vectors and matrices to higher dimensions.
They're very useful.
http://www.dailymail.co.uk/news/article-1369595/Jacob-Barnett-12-higher-IQ-Einstein-develops-theory-relativity.html
my goal, i am currently 11 years behind
It also depends on what areas you want to focus on in maths. There's discrete mathematics/number theory, mathematical logic, topology, algebra, mathematical statistics,...
Yeah, I'be heard of him. I hope you get there (minus the autism).
*I've*
We 'know' general relativity is wrong - feel like fixing it, Arman?
sounds like a plan
What about bed?
Might be the first step.
oiudhvcapoufimnac
Is that code?
fine -.-
lol
nah, gibberish.... alright my chances of catching up to an ausbergers kid is highly unlikely, but i can at least try to match his pace...
i think... idk. i will win -.-
yeah, that's the right attitude
Just ace your doughnuts.
lol iight gnight yo
nite
hey arman
I am becomeMyFan lol, got a new account because I got tired of my old name
hey loki=D
hello
how are you doin?
do you like my new name LOL
ok. bit tired. should be doing my work but wasting time
Yeah, it's cool...did BMF die?
yeah
sad
RIP, BMF
LOL =D, or as sstarica says it, ^_^
Is oktalBlizzard meant to mean something?
i just fanned you
All that work you did collecting fans, gone!
well, it is just my internet user name/ nick name.Oktal means base 8, and blizzard is a natural disaster LOL, thanx I fanned you too, the 132nd fan
=D yeah
As in base 8 for numbers?
but now I am planning to help people more once I learn some more stuff and write my exams
yes, for numbering system, like binary, octal, hexadecimal...
Cool.../random
well, I wanted it to be a bit diferent then my usual nicks
Arman's gone to bed
:)
How's the studying?
great
wow, you're probably the first person i've even asked that question who's said that
:) but really it is great, because I made up my mind to enjoy maths and physics and stuff like this
it is all about the attitude, I think
lol
it is
research backs it
If you assume, when you get to something in maths that you don't understand, that you can understand it eventually, well, the research says those people are more successful.
Took longer to say that than I had anticipated.
Remains of the day.
I just looked at my About me, and noticed something LOL, "...I have infinite interesting ideas but my limit is time..." coincidense but that actually reminds me of something I have learned in maths
lol
Should you be studying versus helping others?
well, I am actually solving questions my self, but once every hour or something I come here to solve a problem or two, to change the mental environment.
lol, okay. you seem sorted.
\[\sum_{\lim_{n \rightarrow \infty}}^{n} n\] where n is ideas something like this? can you improve it?
You could just say\[\left| \left\{ I_{oB} \right\} \right|=\infty\]
The cardinality (size) of the set of oktalBlizzard's ideas is infinite.
:) nice
too bad we can include maths symbols in about me text
You can refer people here to get a better idea.
=D
so, have you checked out the new study groups? Physics, Chemistry and Computer Science?
I had a look at physics yesterday and I was the only one online
me too
I haven't looked at the others.
I don't know anything else about the other subjects. Well, enough to offer any advice.
Q: How does one insult a mathematician? A: You say: "Your brain is smaller than any ε > 0"
lol, that's pretty good.
Theorem: Every positive integer is interesting. Proof: By contradiction, assume that there exists an uninteresting positive integer. Then there must be a smallest uninteresting positive integer. But that's pretty interesting! Therefore a contradiction!
Let epsilon be less than zero...
Are you coming up with this random stuff?
In some alley, a function meets up with a differential operator: "Get out of my way - or I'll differentiate you till you're zero!" "Try it - I'm \[e ^{x} \]..." "Too bad... I'm d/dy." no, I got this from a friend
I have one.
Pi had an argument with i. i told pi to be rational and pi told i to get real
Nice! LOL, I think I heared something like that, Check this: Salary Theorem: The less you know, the more you make. Proof: Fact #1: Knowledge is Power Fact #2: Time is Money We know that: Power = Work / Time And since Knowledge = Power and Time = Money It is therefore true that Knowledge = Work / Money Solving for Money, we get: Money = Work / Knowledge Thus, as Knowledge approaches zero, Money approaches infinity, regardless of the amount of Work done
Yeah, I think that's more sadly true than funny ;p
aiight OB, I have to go. Have to do some work before the days runs out and I feel guilty ;)
he also gave me this, but I dont get it at all, maybe you do: A SLICE OF PI ****************** 3.14159265358979 1640628620899 23172535940 881097566 5432664 09171 036 5 ok, bye, see you next time
hehe..see you :)
lolol

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