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

1. Aug 26, 2005

### Rosnet

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. Aug 26, 2005

### lurflurf

We have
$$\frac{d(xy)}{dx}=a^{xy}$$
multiply both sided by a known derivative
$$\frac{d(a^{-xy})}{d(xy)} \ \frac{d(xy)}{dx}=a^{xy}\frac{d(a^{-xy})}{d(xy)}$$
use the chain rule on the left the know derivative on the right
$$\frac{d(a^{-xy})}{dx}=a^{xy}a^{-xy}\frac{d(-xy)}{d(xy)}\log(a)$$
simplify and integrate
$$a^{-xy}=\int -\log(a)dx$$
you should be able to finish things up

3. Aug 26, 2005

### lurflurf

Another perhaps simpler method
$$\frac{d(xy)}{dx}=a^{xy}$$
by the chain rule
$$\frac{du}{dv} \ \frac{dv}{du}=1$$
so
$$\frac{dx}{d(xy)}=a^{-xy}$$
write in differential form
$$dx=a^{-xy}d(xy)$$
integrate
$$x+C=-\frac{a^{-xy}}{\log(a)}$$
means a^(xy) not (a^x)y

Last edited: Aug 26, 2005
4. Aug 27, 2005

### asdf1

@@a
wow~
how'd you think of those two methods?

5. Aug 27, 2005

### lurflurf

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
6. Aug 28, 2005

### GCT

$$\frac{d(a^{-xy})}{dx}=a^{xy}a^{-xy}\frac{d(-xy)}{d(xy)}\log(a)$$

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

7. Aug 29, 2005

### asdf1

you have a really great mind~
@@a
when i first saw that problem, nothing sprang into my mind...

8. Aug 29, 2005

### lurflurf

It is just the chain rule of single variable calculus.
$$\frac{df}{dx}=\frac{df}{du} \ \frac{du}{dx}$$
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
$$\frac{d(\sin(xy))}{d(xy)}=\cos(xy)$$
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.
$$\frac{d(a^{-xy})}{dx}=-a^{-xy}\log(a)\frac{d(xy)}{dx}=-a^{xy}a^{-xy}\log(a)=-\log(a)$$
where the given differential equation
$$\frac{d(xy)}{dx}=a^{xy}$$
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
$$\frac{d(xy)}{dx}=a^{ax}$$
obvious algebra
$$a^{-xy}\frac{d(xy)}{dx}=1$$
integrate
$$\int a^{-xy}\frac{d(xy)}{dx} \ dx=\int 1 \ dx$$
performing the integration (the left can be easily seen substituting u=xy)
$$\frac{-1}{\log(a)}a^{-xy}=x+c$$
now the same letting u=xy throughout
$$\frac{d(u)}{dx}=a^{u}$$
obvious algebra
$$a^{-u}\frac{d(u)}{dx}=1$$
integrate
$$\int a^{-u}\frac{d(u)}{dx} \ dx=\int 1 \ dx$$
performing the integration
$$\frac{-1}{\log(a)}a^{-u}=x+c$$

9. Aug 29, 2005

### GCT

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. Aug 29, 2005

### lurflurf

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.