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    Can anybody help me solve this:

    Code:
    dy/dx = 1/( e^y - x )
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    i can't see how to do this by one of the standard a level methods -- however, i might be wrong -- but you can get somewhere with it "by inspection".

    look at the right hand side

    for
    1/(e^y - x)
    to take a "nice" form you need e^y to be some multiple of x.

    So assume y=ln(kx) where k is just some number to be worked out and see what happens

    left hand side
    ==========

    y=ln(kx) => dy/dx = k/kx = 1/x

    right hand side
    ===========

    y=ln(kx) => 1/(e^y - x) = 1/(e^(ln(kx)) - x) = 1/(kx - x) = 1/((k-1)x)

    so, if the differential equation is satisfied
    ==============================

    1/x = 1/((k-1)x) => 1 = k -1 => k = 2

    Hence solution
    ===========

    y = ln(2x)

    however
    ======

    the solution should have a constant of integration in it somewhere...i have missed this out in the above...try putting it a few places and see which one cancels on both sides might work...there again maybe not...hopefully the above is some help, tho. the following might help also
    http://www.efunda.com/math/ode/ode1_implicit.cfm
    but maybe a bit advanced?
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    Nice solution. Thanks for the help.


    I think you are right about the constant, it doesn't seem to work for y = ln(2Cx) or y = ln(2x) + C because the left hand side will always be 1/x and the right will only be 1/x for k = 2
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    (Original post by mikesgt2)
    Can anybody help me solve this:

    Code:
    dy/dx = 1/( e^y - x )
    This is easy to solve if we note that dy/dx = 1/(dx/dy) and use this to change the independent variable. Then we use the standard integrating factor technique.

    dy/dx = 1/(e^y - x) <=> dx/dy = e^y - x

    dx/dy = e^y - x
    dx/dy + x = e^y
    (e^y)(dx/dy) + (e^y)x = e^(2y)
    x(e^y) = Integral(e^(2y) dy)
    x(e^y) = (1/2)e^(2y) + C
    x = (1/2)(e^y) + C(e^(-y))

    I hope this helps,
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    Thanks... that is a lot nicer.
 
 
 
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