How Do You Solve the Differential Equation kxy = (y+dx)(x+dy)?

[r16]

I have run into this problem solving differential equations of this type (they occur often doing momentum problems):

$$kxy = (y+dx)(x+dy)$$

where $k$ is constant. I multiply it out to :

$$kxy= xy + xdx + ydy + dydx$$

Regroup and :

[tex] \int {kxy} = \int {xdx} + \int {ydy} + \int {dydx} [/itex]

I'm left with the term $\int dxdy$ that I don't know what to do with. Am I able to hold either the $dx$ or $dy$ constant and integrate with respect to the other? I am not able to find a transformation that will remove the $dydx$ or $\frac{dy}{dx}$ or $\frac{dx}{dy}$. I am also confused about the term $\int kxy$: integration without respect to a particular differential. How would I solve this differential equation?

 

[HallsofIvy]

There something wrong with your equation. You can't have "dx" and "dy" mixed like that. If you are dealing with "differentials" dx and dy, it might make sense (but it would just say kxy= xy) but it is certainly not a differential equation.

 

[tacman]

It looks like you have made good progress in trying to solve this differential equation. The first step, as you have done, is to expand and regroup the terms. However, it seems like you may have made a mistake in your integration steps. Let's take a closer look at your work:

kxy = xy + xdx + ydy + dydx

Next, you integrated each term separately, which gave you:

[tex] \int {kxy} = \int {xy} + \int {xdx} + \int {ydy} + \int {dydx} [/itex]

But the problem is with the last term, [tex] \int {dydx} [/itex]. This is not a valid integral because it contains both dx and dy. Remember, when we integrate with respect to a variable, we treat all other variables as constant. So in this case, we would have:

[tex] \int {dydx} = y + C [/itex]

where C is a constant. This does not help us solve the differential equation.

One possible approach to solving this differential equation is to use the method of separation of variables. This involves separating the variables (in this case, x and y) onto opposite sides of the equation and then integrating both sides separately. This would give us:

[tex] \int {\frac{k}{x+dx}} dx = \int {\frac{1}{y+dy}} dy [/itex]

To solve this, we would need to find a way to simplify the fractions on both sides. It may be helpful to use a substitution, such as u = x+dx and v = y+dy. This would give us:

[tex] \int {\frac{k}{u}} du = \int {\frac{1}{v}} dv [/itex]

Now we can integrate both sides and solve for u and v, which will give us the solution for x and y. Remember to include the constant of integration when integrating both sides.

Another possible approach is to use a substitution to convert the equation into a separable form. For example, we could let u = xy, which would give us:

[tex] \frac{du}{dx} = \frac{y+dx}{x+dy} [/itex]

This is now a separable differential equation, which can be solved using the method of separation of variables.

In summary, solving