Question

help! how do i find the general soln for y'=x/y +y/x

Answer 1

x/y +y/x=xy+yx/xy=2xy/xy=2
Got it ?:)

Answer 2

I want to separate the variables, but I am having a really hard time.

Answer 3

separate the variables ???
What is your mean ???

Answer 4

how did you get 2xy/xy?

Answer 5

Because of to find a think to + them its /xy PK ?:)

Answer 6

OH.

Answer 7

OK means :)

Answer 8

It's hard to see it all in one line. i wrote out what you responded first, and it equals 2. I was expecting an equation of some sort.

Answer 9

so, what do we do with the dy/dx?

Answer 10

Amm ...
Just :

OK ?:)

Answer 11

so dy/dx=2? Don't we integrate both sides?

Answer 12

y' = x/y +y/x
dy/dx = (x^2+y^2)/xy

Answer 13

before we integrate we need to seperate x and y completely at different sides

Answer 14

that's exactly what I thought when I first started. The other person who responded said otherwise.

Answer 15

I need help separating the variables.

Answer 16

I am trying to do that...... looks really difficult for me

Answer 17

yeah, it really is. there must be some simple trick to it.

Answer 18

Did u get it ???

Answer 19

no, i didnt. I am still struggling.

Answer 20

So do u want try again ?:)

Answer 21

yes, can you see what you can figure out? separating the variables is difficult with this particular one.

Answer 22

Sure !
No !They not hard if u know :)

Answer 23

thanks. ill be awaiting your reply.

Answer 24

heh !
Now start with this question :
1/2+2/3=?

Answer 25

?

Answer 26

is this some kind of joke?

Answer 27

OK !
Just answer please :)

Answer 28

7/6

Answer 29

i found a common denominator

Answer 30

OK !
Please say how because we have 2 & 3 and we dont have 2=3 .!

Answer 31

\[y' =\frac{x}{y} +\frac{y}{x}\]\[y' = \frac{x^2 + y^2}{xy}\]this is a first-order homogeneous DE

Answer 32

Thanks exraven. I got up to the second line, and I tried manipulating it around, however, it's just not coming out nicely.

Answer 33

huh !
Got it ????

Answer 34

the method to solve first-order homogeneous DE is by using the substitution \[y = zx\]\[y' = z'x + z\]if we plug it in,\[z'x + z = \frac{x^2 + z^2x^2}{x^2z}\]\[z'x + z = \frac{1 + z^2}{z}\]\[z'x = \frac{1}{z}\]\[\frac{dz}{dx}x = 1/z\]\[z\:dz = \frac{1}{x}dx\]just integrate both sides and substitute back to y and x

Answer 35

what does the "z" represent? a constant?

Answer 36

oh. I think "z" is like the equivalent of "u"

Answer 37

nope, a function

Answer 38

we could also consider it like this\[z = \frac{y}{x}\] so after we plug it in to the DE, z will become the dependent variable

Answer 39

what is that substitution technique called? I just finished following through all the algebra and it makes perfect sense but the z part. it still confuses me.

Answer 40

use pauls online notes website, I used that when I was self studying calculus and diff eq, here

http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx

Answer 41

so from my understanding, that substitution technique can be used if it is a homogeneous DE, right? does it have to be a 1st order?

Answer 42

yes, that technique only applies if and only if the DE is first-order and homogeneous

Answer 43

thanks for citing pauls online notes. I've used his notes before and theyre good. although what you linked me to... well... the concepts are just pretty dense. I'm still digesting it all.

Answer 44

but in layman's terms, can you explain that substitution rule just a little bit more in your own words? thanks so much for your time and effort. I'm also really impressed with the typed out work you posted earlier. how'd you do that?

Answer 45

\[y' = \frac{x}{y} + \frac{y}{x}\]\[y' = \frac{1}{\frac{y}{x}} + \frac{y}{x}\]by inspection, we see that the DE is a first-order DE, and because we can write the DE as\[y' = F\left(\frac{y}{x} \right)\]the DE is homogeneous, therefore we can apply the substitution \[z = \frac{y}{x}\]rearrange and apply the chain rule to get the derivative\[y = zx\]\[y = z'x + z\]

Answer 46

hey exraven, so when integrated, i get 1/2z^2 = lnx +c. Do I also need a +c on the left side?

Answer 47

nope, you just need to add the constant either on the left or the right side

Answer 48

and so is my final general soln is:

(1/2)(x/y)^2 = |lnx| +c?

Answer 49

Thank you so much for everything so far. I really appreciate it! You're answers are thorough and you've been very helpful. I have another question DE question, and I hope you could help me out with that one too.

Answer 50

sorry I have to go now, I am busy irl

Answer 51

ok np. but is my general soln correct?

Answer 52

\[\frac{1}{2}\left(\frac{y}{x}\right)^2 = \ln |x| + C\]

Answer 53

just saw this just now. Thanks!