Question

just want help separating like terms in the following:
xy + y' = 100x

Answer 1

factor right?

Answer 2

\[y'=x(100-y)\]

Answer 3

divide both sides by 100-y

Answer 4

ohh ok thakns

Answer 5

Pro challenge: evaluate y

Answer 6

\[y'=100x-xy\]
\[y'=x(100-y)\]
\[\frac{y'}{100-y}=x\]

Answer 7

i actually don't know this, so don't trust me, trust myininaya

Answer 8

\[\int\limits_{}^{} \frac{dy}{100-y}=\int\limits_{}^{} x dx\]

Answer 9

it seems like you do satellie

Answer 10

wow
no actaully i was shooting from the hip
it did say "separate" and i can read, but i know next to nothing about diffeq

Answer 11

haha well thanks for the help guys. i can take it from here, but feel free to try anything you want

Answer 12

let me guess, now we get
\[\frac{x^2}{2}+c=\log(\text{something})\] and then exponentiate right?

Answer 13

yep -log(something)

Answer 14

or yeah -log(something)=log(something^(-1)) lol

Answer 15

so either way lol

Answer 16

something^(-1) is still something

Answer 17

not the same something of course

Answer 18

oh it is just
\[\log(100-y)=\frac{x^2}{2}+c\]
\[100-y=e^{\frac{x^2}{2}+c}\] how am i doing so far? then
\[y=100-e^{\frac{x^2}{2}+c}\] close?

Answer 19

aren't you missing something satellite in the first line

Answer 20

I really meant that challenge for the question poster, but this works too, lols.

Answer 21

could be help a brother out

Answer 22

The negative, mmmyeesss

Answer 23

I got the same as satellite only with a plus not a minus...

Answer 24

yeah @creative, ignore me i am just messing around

Answer 25

Satellites don't mess around, they orbit stuff.

Answer 26

too many details to keep track of

Answer 27

\[-\log(100-y)=\frac{x^2}{2}+c\] that better?

Answer 28

well how long has it been since you guys took calc? probably gives enough leeway to be *allowed* to forget

Answer 29

:P

Answer 30

when i took math calc hadn't been invented yet

Answer 31

lol

Answer 32

When I took calc, satellites weren't invented yet.

Answer 33

even natural satellites? O.o
kidding :)

Answer 34

just in case you're still curious satellite, this is what I did:
dy/(100-y) = x dx

integrate both sides to get:
ln |100-y| = .5x^2 + C

which after being raised to e is the same as:
|100-y| = |y-100| = e^(.5x^2) * e^C = Ce^(.5x^2)

so the answer: y = Ce^(.5x^2) + 100