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

  • Thread starter Rosnet
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  • #1
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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.
 

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  • #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
 
  • #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
 
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  • #4
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@@a
wow~
how'd you think of those two methods?
 
  • #5
lurflurf
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asdf1 said:
@@a
wow~
how'd you think of those two methods?
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).
 
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  • #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?
 
  • #7
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you have a really great mind~
@@a
when i first saw that problem, nothing sprang into my mind...
 
  • #8
lurflurf
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GCT said:
[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?
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]
 
  • #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.
 
  • #10
lurflurf
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GCT said:
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.
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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