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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
it is a fairly drawn out process... have you checked out http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx
that website has the answer do your exact question, just follow the steps
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.
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
Did you get it?
hmm so I have (1/e^v) dv = (1/x) dx now
then do I just try and solve this new thing like a seperable differential equation?
integrate both sides integral of e^-x dx is -e^-x and integral of 1/x is ln x
so now I have -e^-v = ln|x| + C
now I can just subsitute y/x back in and take the ln to isolate the y?
v=y/x so substitue back to get the equation in terms of y and x.
ok so I have (y/x)*ln|-e| = ln|(ln|x| + C)| then y = x*ln|(ln|x| + C)| ------------- ln|e|
is this correct?
ln e=1. yeah it is correct.
so just y = x*ln|(ln|x| + C)|
Hold on Theres a negative sign too. it is e^(-y/x). So taking ln leaves a minus sign too.
oops then y = -x*ln|(ln|x| + C)|
yeah this shld be rite.
alright thanks a lot!