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anonymous

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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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

  2. anonymous
    • 4 years ago
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    it is a fairly drawn out process... have you checked out http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx

  3. anonymous
    • 4 years ago
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    that website has the answer do your exact question, just follow the steps

  4. anonymous
    • 4 years ago
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    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.

  5. anonymous
    • 4 years ago
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    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

  6. anonymous
    • 4 years ago
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    oh

  7. anonymous
    • 4 years ago
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    Did you get it?

  8. anonymous
    • 4 years ago
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    hmm so I have (1/e^v) dv = (1/x) dx now

  9. anonymous
    • 4 years ago
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    then do I just try and solve this new thing like a seperable differential equation?

  10. anonymous
    • 4 years ago
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    integrate both sides integral of e^-x dx is -e^-x and integral of 1/x is ln x

  11. anonymous
    • 4 years ago
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    so now I have -e^-v = ln|x| + C

  12. anonymous
    • 4 years ago
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    now I can just subsitute y/x back in and take the ln to isolate the y?

  13. anonymous
    • 4 years ago
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    v=y/x so substitue back to get the equation in terms of y and x.

  14. anonymous
    • 4 years ago
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    yeah correct.

  15. anonymous
    • 4 years ago
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    ok so I have (y/x)*ln|-e| = ln|(ln|x| + C)| then y = x*ln|(ln|x| + C)| ------------- ln|e|

  16. anonymous
    • 4 years ago
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    is this correct?

  17. anonymous
    • 4 years ago
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    ln e=1. yeah it is correct.

  18. anonymous
    • 4 years ago
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    oh right

  19. anonymous
    • 4 years ago
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    so just y = x*ln|(ln|x| + C)|

  20. anonymous
    • 4 years ago
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    Hold on Theres a negative sign too. it is e^(-y/x). So taking ln leaves a minus sign too.

  21. anonymous
    • 4 years ago
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    oops then y = -x*ln|(ln|x| + C)|

  22. anonymous
    • 4 years ago
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    yeah this shld be rite.

  23. anonymous
    • 4 years ago
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    alright thanks a lot!

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