Question

Can someone help me solve this differential equation?
dy/dx = y/x + e^(y/x)

Answer 1

so far what I've done is subsitute v = y/x
I'm just haveing some trouble with what to do with the dy/dx after the subsitution

Answer 2

it is a fairly drawn out process... have you checked out
http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx

Answer 3

that website has the answer do your exact question, just follow the steps

Answer 4

After you put y=vx. you get
dy/dx= v+ xdv/dx.
dy/dx=v+e^v.
v cancels you get
e^v=xdv/dx.
e^-v dv=dx/x.
e^-v+log x=c.

Answer 5

so as far as I can tell you islate y
y=vx
then take the derivative in terms of x on both sides to get
dy/dx = (dv/dx)x + v

Answer 6

oh

Answer 7

Did you get it?

Answer 8

hmm so I have
(1/e^v) dv = (1/x) dx now

Answer 9

then do I just try and solve this new thing like a seperable differential equation?

Answer 10

integrate both sides integral of e^-x dx is -e^-x and integral of 1/x is ln x

Answer 11

so now I have -e^-v = ln|x| + C

Answer 12

now I can just subsitute y/x back in and take the ln to isolate the y?

Answer 13

v=y/x so substitue back to get the equation in terms of y and x.

Answer 14

yeah correct.

Answer 15

ok so I have (y/x)*ln|-e| = ln|(ln|x| + C)|
then
y = x*ln|(ln|x| + C)|
-------------
ln|e|

Answer 16

is this correct?

Answer 17

ln e=1. yeah it is correct.

Answer 18

oh right

Answer 19

so just y = x*ln|(ln|x| + C)|

Answer 20

Hold on Theres a negative sign too. it is e^(-y/x). So taking ln leaves a minus sign too.

Answer 21

oops then y = -x*ln|(ln|x| + C)|

Answer 22

yeah this shld be rite.

Answer 23

alright thanks a lot!