Question

FODQ, solve for general solution
(x-y)y'=x+y

y' =(x+y)/(x-y)
y'/x = (1+y/x)/(1-y/x)

let v = y/x

v/x + v' = (1+v)/(1-v)

it appears I can't cleanly seperate these variables, but this is clearly homogeneous..

Answer 1

y = 2tan(y/x)^-1 - ln(y^2/x^2 + 1) = 2ln(x) + c

Answer 2

good way to test as to is it homogeneous....given f' = (x,y), then f(x, xt) must equal f(1,t)

Answer 3

never heard of such a test, I'm interested. It seemed most likely in this form compared to direct or bernoulli

Answer 4

all homogeneous first order d.eq's are separable after v=y/x substi.

Answer 5

so I was making lemonade out of charcoal? then this must be direct? I've attemped direct and could't seperate them either

Answer 6

ayan -try my cute definition for homogeneous 1st orders

Answer 7

v =y/x vx=y y'= v+v'x

Answer 8

correct

Answer 9

i don't follow however, i've used those subsitutions, and ended up with an unseperable equation

Answer 10

all homogenous 1st orders can be separated

Answer 11

am I utilizing the wrong subsitution?

Answer 12

I can't see bernoulli will help me in this situation.

Answer 13

bernoulli pf of the fork y'+ f(x)y = g(x) * y^n where n cannot = 0 or 1 pf should read is...fork reads form

Answer 14

i teach diff eq. using boyce and diprima's brilliant book

Answer 15

thank you, I haven't solved this yet, but i'll try hacking at it my putting it in bernoulli form. Given the daunting final result, I'm not sure if I'll calculate it correctly.

Answer 16

what is the problem?

Answer 17

hacking into it again now.

Answer 18

your sub is wrong...the left side should be v + V'x

Answer 19

y' = v + v'x since i have y'/x this equates to v/x + v' does it not?

Answer 20

you have y' = (x+y)/(x-y) = v+v'x.....times the left by i/x over 1/x, and iys homogeneous and ready for its algebra helper

Answer 21

(1+v)/ (1-v) - v =v'x...that separates

Answer 22

where v' is dv/dx

Answer 23

after cleaning up the rational expression on the left

Answer 24

I'm probably viewing my problem thought a kolidascope, but I don't see how you disposed of y'/x = (1+y/x)/(1-y/x) prior to subsitution

Answer 25

where both sides needed to be divided by x, sorry if I'm sounding imbecile

Answer 26

from there, my left side after subsitution becomes (v+v'x)/x

Answer 27

ur prob was y' = (x+y)/(x-y)

y' = v+v'x = (1+v)/(1-v)....subtract v from both sides and get a common denominator
so, (1+v)/1-v) - v/1= 1+v -(1-v)*v/1 all over 1-v

Answer 28

(1+v-v+v^2) = 1+v^2 / 1-v = v'x

Answer 29

(1-v)dv / (1+v^2) =dx/x

Answer 30

i'm still rolling around the first statement in my head, I under stand how you proceed from
y'=v+v'x = (1+v)/(1-v) it's just that to get that v sub in, I still believe that y'/x had to have occured. Why do I believe that I wonder

Answer 31

v=y/x.......to get y'. you need y = vx

Answer 32

then product rule with vx

Answer 33

yes, y' = x'v + v'x
and x' = 1 so v'=v + v'x

I'm shaking the stupid from my head but this nugget is stuborn

Answer 34

but i have only taught this stuff for 20 yrs :)

Answer 35

no no y' = v +v'x

Answer 36

I must be a vast disapointmet to encounter then by your reckoning..

Answer 37

yes y', sorry, typo on that occurance

Answer 38

not at all...we are all tiny players

Answer 39

do you see how it separates now?

Answer 40

if i could do it on a board for you, it would take 2 minutes

Answer 41

yes but I still have that nugget, I think I'm thinking something wrong. In all honesty, the integral is supposed to be harder than seperating, I have no idea why this bump is becoming a boulder

Answer 42

all homo first orders can be separated

Answer 43

so I see it, I'm thinking, that i'm just doing a mathematical error similar to losing a negative of sorts involving 1/x

Answer 44

perhaps I should state my problem clearly

Answer 45

sure

Answer 46

\[y' = (x+y)/(x-y)\]
mult 1/x on LHS and RHS
\[y'/x = (x/x + y/x)/(x/x - y/x) = (1+ y/x)(1-y/x)\]
let v = y/x
so
\[y'/x = (1+v)/(1-v)\] = \[(v+v'x)/x = (1+v)/(1-v)\]

Answer 47

i suppose you can see my rouge 1/x , probably a product of my stupidity

Answer 48

after this however, I can cleanly integrate it, I suppose I'm just not doing my basic fractions right

Answer 49

you only times the top nad bottom by 1/x not the other side....y' reamins as is v + v'x

Answer 50

oh, yes.. and it doesn't change the equation as it's anagolous to rewriting the equation

Answer 51

I feel very stupid now

Answer 52

v+v'x = (1+v)/(1-v)

Answer 53

yes, I see how to proceed, how did I let that slip. Thank you for letting me waste your time, and I appologize for that. Apparently I'm losing grip with my sanity.