anonymous
  • anonymous
Can someone help me solve this differential equation? dy/dx = y/x + e^(y/x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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
anonymous
  • anonymous
it is a fairly drawn out process... have you checked out http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx
anonymous
  • anonymous
that website has the answer do your exact question, just follow the steps

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anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
oh
anonymous
  • anonymous
Did you get it?
anonymous
  • anonymous
hmm so I have (1/e^v) dv = (1/x) dx now
anonymous
  • anonymous
then do I just try and solve this new thing like a seperable differential equation?
anonymous
  • anonymous
integrate both sides integral of e^-x dx is -e^-x and integral of 1/x is ln x
anonymous
  • anonymous
so now I have -e^-v = ln|x| + C
anonymous
  • anonymous
now I can just subsitute y/x back in and take the ln to isolate the y?
anonymous
  • anonymous
v=y/x so substitue back to get the equation in terms of y and x.
anonymous
  • anonymous
yeah correct.
anonymous
  • anonymous
ok so I have (y/x)*ln|-e| = ln|(ln|x| + C)| then y = x*ln|(ln|x| + C)| ------------- ln|e|
anonymous
  • anonymous
is this correct?
anonymous
  • anonymous
ln e=1. yeah it is correct.
anonymous
  • anonymous
oh right
anonymous
  • anonymous
so just y = x*ln|(ln|x| + C)|
anonymous
  • anonymous
Hold on Theres a negative sign too. it is e^(-y/x). So taking ln leaves a minus sign too.
anonymous
  • anonymous
oops then y = -x*ln|(ln|x| + C)|
anonymous
  • anonymous
yeah this shld be rite.
anonymous
  • anonymous
alright thanks a lot!

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