Question

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)

Answer 1

initial condition is (0,-2)

Answer 2

You can use separation by variables here and then integrate. Once you have y after integration, you can use the initial condition.

Answer 3

i did that, but my solution is y=(-2/x^2)+2
which doesnt technically do through the initial condition

Answer 4

i think its a vertical asymptote which... definitely isint the point XD

Answer 5

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,\[0-2c+1=0 \rightarrow c=\frac{1}{2}\]

Answer 6

You can then factor out y and rearrange into the form y=y(x).

Answer 7

brb

Answer 8

i got c=2?

Answer 9

When you separated the variable and tried to get y by itself did you include c and also multiply the y by it?

Answer 10

.... -.-

Answer 11

ima try that, how about: y'=1+yy

Answer 12

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.

Answer 13

dang, the small things get by me... that should be something like arctany=x+c? so tan(x+c)=y?

Answer 14

that explains why my condition is in radians

Answer 15

Yes.

Answer 16

alright... major confusion...
\[y'=(e ^{x-y})\div(1+e ^{x})\]

Answer 17

You want to solve that?

Answer 18

sadly ><

Answer 19

Just pull out the e^(-y) and it becomes separable ;)

Answer 20

asoghaqoeraifoiafauirfhlanihusdfvnnfldanufd

Answer 21

\[e^ydy=\frac{e^x}{1+e^x}dx\]

Answer 22

differential equations are gay -.-

Answer 23

lol

Answer 24

dont tell me any more!!!

Answer 25

tell you what? answers?

Answer 26

yeah.. dont tell me any more lol

Answer 27

no worries :D

Answer 28

I think you have the best icon...if that helps...

Answer 29

hahahaha it was annoyingly hard to make XD

Answer 30

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

Answer 31

it could be me, since i look like that

Answer 32

lol, how did your frigging doughnut go?

Answer 33

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

Answer 34

alright im stuck at e^y=ln(1+e^x)+c
initial condition is (0,1)

Answer 35

Is that y supposed to be y' (y-prime)?

Answer 36

nope, thats my solution

Answer 37

to the above equation

Answer 38

oh..

Answer 39

Yeah. You're right. Leave it like that too.

Answer 40

but i need to find c

Answer 41

Oh, just plug your numbers in.

Answer 42

\[e=\ln 2 +c \rightarrow c = e- \ln 2\]

Answer 43

any way to do that without a calc?

Answer 44

You mean simplify?

Answer 45

No.

Answer 46

What's wrong with Euler's number? :p

Answer 47

ah god its 1... can you check these real quick?

Answer 48

Check what?

Answer 49

Do you ever sleep?

Answer 50

not too much anymore lol

Answer 51

I need paper...I've run out...hang on...

Answer 52

but all my work is shown?

Answer 53

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.

Answer 54

do i have to multiply the c by 2 aswell?

Answer 55

I'll do it here:

Answer 56

my answer doesnt make sense anyways... when i find y' its all weird and quotienty

Answer 57

\[\frac{dy}{dx}=xy^2 \rightarrow y^{-2}dy = x dx \rightarrow -y^{-1}=x+c\]

Answer 58

Your boundary condition is (0,-2), so

Answer 59

x+c or (1/2)xx+c?

Answer 60

\[-2^{-1}=0+c \rightarrow c=-\frac{1}{2}\]

Answer 61

So\[y^{-1}+x=\frac{1}{2} \rightarrow 2+2xy=y \rightarrow y(1-2x)=2 \rightarrow y=\frac{2}{1-2x}\]

Answer 62

waitttt the integral of xdx is (1/2)xx+c though!

Answer 63

Oh crap, I forgot the frigging x. Sorry Arman.

Answer 64

nah dont be sorry, i should hit the sheets tho, i'll probably have more questions tomorrow lol

Answer 65

I still get a different answer to you.

Answer 66

basic differential equations (like these) arent usually part of calc II are they?

Answer 67

Yours is close. I'll work through them - they won't take long - and scan and you can read it in the morning.

Answer 68

different answer?

Answer 69

I'm not sure about the American system.

Answer 70

y=2/(1-x^2)

Answer 71

did you use (0,-2) for the initial whatever?

Answer 72

pellet, I wrote it wrong on my paper. I wrote (0,2). SORRY AGAIN :'(

Answer 73

NOW I get what you got.

Answer 74

>< nah man i hate when ppl tell me sorry

Answer 75

lol XD

Answer 76

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

Answer 77

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).

Answer 78

thats probably better in the math world, where you need to be precise

Answer 79

Yeah, exactly.

Answer 80

is that prefered in college?

Answer 81

preferred*

Answer 82

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.

Answer 83

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)

Answer 84

?

Answer 85

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.

Answer 86

oh... well what are the chances that i can attain a college grad level (in math) of mathematical understanding in the next year?

Answer 87

given the rate that i covered calc 1 and 2 within 1.5 months

Answer 88

I'd say you're pretty good.

Answer 89

College grad level in calculus...is that what you're asking?

Answer 90

You have to cover calculus of a complex variable as well. There's also tensor calculus.

Answer 91

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?

Answer 92

complex variable wont be too hard, what is tensor calc??

Answer 93

It *might* take a little longer ;p

Answer 94

They're devices we use to extend scalars, vectors and matrices to higher dimensions.

Answer 95

They're very useful.

Answer 96

http://www.dailymail.co.uk/news/article-1369595/Jacob-Barnett-12-higher-IQ-Einstein-develops-theory-relativity.html