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Solving this equation: d(xy)/dx = a^xy

  1. Aug 26, 2005 #1
    Can anyone suggest a method of solving this equation:

    d(xy)/dx = a^xy (Where ^ means raised to the power)

    Don't try a series solution.
     
  2. jcsd
  3. Aug 26, 2005 #2

    lurflurf

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    We have
    [tex]\frac{d(xy)}{dx}=a^{xy}[/tex]
    multiply both sided by a known derivative
    [tex]\frac{d(a^{-xy})}{d(xy)} \ \frac{d(xy)}{dx}=a^{xy}\frac{d(a^{-xy})}{d(xy)}[/tex]
    use the chain rule on the left the know derivative on the right
    [tex]\frac{d(a^{-xy})}{dx}=a^{xy}a^{-xy}\frac{d(-xy)}{d(xy)}\log(a)[/tex]
    simplify and integrate
    [tex]a^{-xy}=\int -\log(a)dx[/tex]
    you should be able to finish things up
     
  4. Aug 26, 2005 #3

    lurflurf

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    Another perhaps simpler method
    [tex]\frac{d(xy)}{dx}=a^{xy}[/tex]
    by the chain rule
    [tex]\frac{du}{dv} \ \frac{dv}{du}=1[/tex]
    so
    [tex]\frac{dx}{d(xy)}=a^{-xy}[/tex]
    write in differential form
    [tex]dx=a^{-xy}d(xy)[/tex]
    integrate
    [tex]x+C=-\frac{a^{-xy}}{\log(a)}[/tex]
    I hope your ambiguous a^xy
    means a^(xy) not (a^x)y
     
    Last edited: Aug 26, 2005
  5. Aug 27, 2005 #4
    @@a
    wow~
    how'd you think of those two methods?
     
  6. Aug 27, 2005 #5

    lurflurf

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    They just spring to mind. They also use the same ideas.
    For the first one I knew I wanted to multiply the right side by a^(-xy) to get a constant, but I wanted it in a form d(f)/d(xy) so that I could use the chain rule on the left, so d(a^(-xy))/d(xy) was a good choice.
    For the second one using the chain rule to flip the derivative seemed to be helpful, since the right hand side has a simple multiplicative inverse, and is a function of xy that is easy to integrate (with respect xy).
     
    Last edited: Aug 27, 2005
  7. Aug 28, 2005 #6

    GCT

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    [tex]\frac{d(a^{-xy})}{dx}=a^{xy}a^{-xy}\frac{d(-xy)}{d(xy)}\log(a)[/tex]

    how'd you take a derivative of two variables, is this multivariable calculus?
     
  8. Aug 29, 2005 #7
    you have a really great mind~
    @@a
    when i first saw that problem, nothing sprang into my mind...
     
  9. Aug 29, 2005 #8

    lurflurf

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    It is just the chain rule of single variable calculus.
    [tex]\frac{df}{dx}=\frac{df}{du} \ \frac{du}{dx}[/tex]
    So we can differentiate with respect to xy or some other function of x. What ever function we use is u in the above. For example
    [tex]\frac{d(\sin(xy))}{d(xy)}=\cos(xy)[/tex]
    The xy can be treated as a variable. It might be easier to see when xy is writen u=xy.
    The equation you quote is only true of y that satisfy the differential equation, but it is easy to see its truth in those cases.
    [tex]\frac{d(a^{-xy})}{dx}=-a^{-xy}\log(a)\frac{d(xy)}{dx}=-a^{xy}a^{-xy}\log(a)=-\log(a)[/tex]
    where the given differential equation
    [tex]\frac{d(xy)}{dx}=a^{xy}[/tex]
    has been used
    I will do this once more, attempting to avoid writing derivatives that look strange. All three of these workings are essentially the same.
    The given equation
    [tex]\frac{d(xy)}{dx}=a^{ax}[/tex]
    obvious algebra
    [tex]a^{-xy}\frac{d(xy)}{dx}=1[/tex]
    integrate
    [tex]\int a^{-xy}\frac{d(xy)}{dx} \ dx=\int 1 \ dx[/tex]
    performing the integration (the left can be easily seen substituting u=xy)
    [tex]\frac{-1}{\log(a)}a^{-xy}=x+c[/tex]
    now the same letting u=xy throughout
    [tex]\frac{d(u)}{dx}=a^{u}[/tex]
    obvious algebra
    [tex]a^{-u}\frac{d(u)}{dx}=1[/tex]
    integrate
    [tex]\int a^{-u}\frac{d(u)}{dx} \ dx=\int 1 \ dx[/tex]
    performing the integration
    [tex]\frac{-1}{\log(a)}a^{-u}=x+c[/tex]
     
  10. Aug 29, 2005 #9

    GCT

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    hmm, I thought that differentiating/integrating two dummy variables might be an issue, although I have only taken integral calculus and intro differential equations, skipped multivariable. Anyways, thanks for the details.
     
  11. Aug 29, 2005 #10

    lurflurf

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    We need not think of it that way since y=y(x)
    In other words xy is some function of x, not a function of x and y.
     
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