Animated Solution for Mathematics - Differential Equations: Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{dp(t)}{dt}=\frac{1}{2}p(t)-200$. If $p(0)=100$, then $p(t)$ equals:

Imagine you are a biologist studying a peculiar colony of rabbits. You have observed that their population growth isn't just a simple exponential curve; it is governed by a specific differential equation:

At first glance, this might look like just another math problem, but it is actually a story of balance. The term $\frac{1}{2}p(t)$ represents the natural tendency of the population to grow, while the $-200$ represents a constant drain—perhaps due to limited resources or predators.

Today, we are going to solve this equation to predict the future of this colony.

The Anatomy of the Equation

We start with our given equation: $\frac{dp(t)}{dt}=\frac{1}{2}p(t)-200$. Our goal is to find $p(t)$, the population at any time $t$, given the initial condition $p(0)=100$.

This is a first-order linear differential equation. To make our lives easier, let us factor out the $\frac{1}{2}$ on the right side to reveal the "equilibrium point" of the system:

Notice something fascinating here? If the population $p$ were exactly $400$, the derivative $\frac{dp}{dt}$ would be zero, and the population would be perfectly stable.

But our rabbits start at $100$, which is well below this equilibrium. This tells us that the population is in a state of decline.

The Dance of Variables

To solve this, we use the method of separation of variables. We want all the $p$ terms on one side and all the $t$ terms on the other.

We rearrange our equation as follows:

This is the moment where the algebra becomes elegant. We have successfully isolated the variables and are now prepared to integrate both sides.

The Integration & The Modulus Trap

Integrating both sides, we get:

Applying the standard integral of $\frac{1}{u}du$, we arrive at:

Here is where most students stumble. We have a modulus sign $|p-400|$. We must determine its sign.

We know $p(0)=100$. Since $100<400$, and the derivative is negative, the population will continue to decrease, meaning $p(t)$ will always be less than $400$.

Therefore, $(p-400)$ is always negative. To remove the modulus, we must multiply by a negative sign: $|p-400|=-(p-400)=400-p$.

Now, let us find the constant $C$ using our initial condition $p(0)=100$:

The Final Synthesis

Substituting $C$ back into our equation, we have:

We want to isolate $p$. Let us move the $\ln 300$ to the left side:

Using the logarithmic property $\ln A-\ln B=\ln (\frac{A}{B})$, we combine the terms:

Finally, we exponentiate both sides to eliminate the natural log:

Multiplying by $300$ and rearranging for $p$, we reach our destination:

We have found the function $p(t)=400-300e^{t/2}$. As time $t$ increases, the term $e^{t/2}$ grows, meaning we are subtracting a larger and larger value from $400$.

This confirms our earlier intuition: the population is declining. Mathematics has allowed us to peer into the future of this rabbit colony with absolute precision.

Remember, the key to these problems is not just the calculation, but understanding the physical reality behind the variables.