- #1

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

Don't try a series solution.

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- Thread starter Rosnet
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- #1

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

Don't try a series solution.

- #2

lurflurf

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[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

[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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@@a

wow~

how'd you think of those two methods?

wow~

how'd you think of those two methods?

- #5

lurflurf

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They just spring to mind. They also use the same ideas.asdf1 said:@@a

wow~

how'd you think of those two methods?

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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how'd you take a derivative of two variables, is this multivariable calculus?

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you have a really great mind~

@@a

when i first saw that problem, nothing sprang into my mind...

@@a

when i first saw that problem, nothing sprang into my mind...

- #8

lurflurf

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It is just the chain rule of single variable calculus.GCT said:

how'd you take a derivative of two variables, is this multivariable 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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- #10

lurflurf

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We need not think of it that way since y=y(x)GCT said:

In other words xy is some function of x, not a function of x and y.

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