Math  /  Calculus

Questiondydt=0.17y(1y400)y(0)=5\begin{array}{l}\frac{d y}{d t}=0.17 y\left(1-\frac{y}{400}\right) \\ y(0)=5\end{array}

Studdy Solution

STEP 1

What is this asking? We're looking at a population of something, maybe cute bunnies, growing over time, and we want to figure out how many bunnies we'll have after a certain amount of time, knowing how their population changes and how many we started with. Watch out! Don't mix up the **initial population** with the **rate of growth**!
Also, remember that the carrying capacity limits how big the population can get!

STEP 2

1. Recognize the Logistic Growth Model
2. Solve for y(t)y(t)

STEP 3

Hey everyone!
We've got this equation, dydt=0.17y(1y400)\frac{dy}{dt} = 0.17y\left(1 - \frac{y}{400}\right), which is a **logistic growth model**!
This type of equation describes how populations grow when there's a limit to how big they can get.
It's like a bunny farm where there's only enough space and food for a certain number of bunnies.

STEP 4

Here, yy is the **population** at time tt, and dydt\frac{dy}{dt} tells us how fast the population is changing.
The **growth rate** is 0.170.17, meaning the population would grow by 17% if there were no limits.
The **carrying capacity**, represented by 400400, is the maximum population the environment can support.
It's like the bunny farm can only hold **400** bunnies max!

STEP 5

We also know the **initial population**, y(0)=5y(0) = 5.
We're starting with just **five** bunnies!

STEP 6

The general solution to a logistic growth equation is y(t)=K1+Aerty(t) = \frac{K}{1 + Ae^{-rt}}, where KK is the **carrying capacity**, rr is the **growth rate**, and AA is a constant we need to figure out.

STEP 7

In our case, K=400K = 400 and r=0.17r = 0.17, so our equation becomes y(t)=4001+Ae0.17ty(t) = \frac{400}{1 + Ae^{-0.17t}}.

STEP 8

To find AA, we use our **initial condition**, y(0)=5y(0) = 5.
Plugging in t=0t = 0, we get 5=4001+A5 = \frac{400}{1 + A}.
Now, we solve for AA: multiply both sides by 1+A1 + A to get 5(1+A)=4005(1 + A) = 400, which simplifies to 5+5A=4005 + 5A = 400.
Subtracting 55 from both sides gives 5A=3955A = 395.
Finally, dividing by 55 gives us A=79A = 79.

STEP 9

Now that we have AA, we can write the particular solution for our bunny population: y(t)=4001+79e0.17ty(t) = \frac{400}{1 + 79e^{-0.17t}}.
This equation tells us how many bunnies we'll have at any given time!

STEP 10

The solution to the differential equation, given the initial condition, is y(t)=4001+79e0.17ty(t) = \frac{400}{1 + 79e^{-0.17t}}.

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