You have the differential equation:

$\frac{d(p)}{d(t)} = \frac{-2k p M}{1 + m k t}$

where $p = p(t)$ is a function of $t$, and $k$, $M$, and $m$ are constants.

This is a first-order linear differential equation that involves the variable $p$ and the independent variable $t$.

To solve this differential equation:

1. Rearrange the equation to isolate the terms involving $p$ on one side:

$\frac{1}{p} \frac{d(p)}{d(t)} = \frac{-2k M}{(1 + m k t)}$

2. Integrate both sides:

   On the left side, we have the derivative of $\ln(p)$ with respect to $t$, since:

$\frac{d}{d(t)} \left( \ln(p) \right) = \frac{1}{p} \frac{d(p)}{d(t)}$

Thus, the equation becomes:

$\frac{d}{d(t)} \left( \ln(p) \right) = \frac{-2k M}{(1 + m k t)}$

Now, integrate both sides:

• Left side:

$\int \frac{d}{d(t)} \left( \ln(p) \right) dt = \ln(p)$

• Right side:

$\int \frac{-2k M}{(1 + m k t)} dt$

Use a simple substitution to solve the right side. Let:

$u = 1 + m k t \quad \Rightarrow \quad du = m k dt$

So, the integral becomes:

$\int \frac{-2k M}{u} \cdot \frac{du}{mk} = -2M \ln(u)$

Substitute back $u = 1 + m k t$:

$-2M \ln(1 + m k t)$

3. Put the result together:

After integration, the solution to the differential equation is:

$\ln(p) = -2M \ln(1 + m k t) + C$

where $C$ is the constant of integration.

4. Solve for $p$:

Exponentiate both sides to get rid of the logarithm:

$p(t) = e^{C} \cdot (1 + m k t)^{-2M}$

Let $e^C$ be a new constant, say $C_1$, then:

$p(t) = C_1 (1 + m k t)^{-2M}$

This is the general solution to the differential equation.

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Would you like me to explain any specific step in more detail, or is there something else you'd like to know? Here are some related questions:

1. What does this solution represent in terms of physical quantities (if any)?
2. How would you find the constant $C_1$ given an initial condition?
3. What is the behavior of $p(t)$ as $t \to \infty$?
4. How can we apply this solution to model population growth or decay problems?
5. Can this solution be generalized for different types of differential equations?

Tip: To solve differential equations like this, it’s often helpful to recognize that it’s separable. Once you rearrange terms properly, the solution involves simple integration steps.