Differential Equations - Exponential Decay

• stanford1463
In summary: For the first interval, where y0 < R/k, the exponential term is decreasing, and the solution is a straight line along R/k.For the second interval, where y0 = R/k, the exponential term is constant, and the solution is at R/k.For the third interval, where y0 > R/k, the exponential term is increasing, and the solution is at a maximum at R/k.
stanford1463

Homework Statement

We have the ODE y' = -ky + R for a population y(t) where death rate exceeds birth rate, counteracted by a constant restocking rate.
I'm assuming k is the decay constant and R is the restocking rate

The population at time t0 = 0 is y0, and I have to find a formula for y(t)
Also, interpret the solution in terms of the long term behavior (0< y0 < R/k, y0 = R/k, and y0 > R/k).

y' = -ky + R

The Attempt at a Solution

I solved the DE and got y = R/k + Ce^(-kt)
But how do I know what this does in those given intervals of y0?
Also if k = .5, R = 2, how would I graph the solutions?

You should get three different graphs for your solution, when you substitute y(0) = y0 into the solution you found. One solution is for y0 between 0 and R/k, one is for y0 = R/k, and one is for y0 > R/k.

Think about these in terms of replenishment rate < death rate, replenishment rate = death rate, and replenishment rate > death rate. How is the population affected in each case?

So...first interval - exponentially decreasing graph
when y = R/k, it's a constant, straight line along R/k?
and when y is big, it's increasing?

Sounds reasonable. Are these what you think, or does your solution back them up?

These are what I think
I'm not sure how to find graph solutions from my equation: y = 4 + Ce^(-.5t)

Thanks for your help!

Where do the k=.5 and R=2 come from? Are they part of the problem statement? You didn't mention these values there. I'm going to ignore them for the time being.

You are given that y(0) = y0, so substitute this into your solution equation to find C.
y(t) = R/k + Ce-kt
y(0) = y0

==> y0 = R/k + C
==> C = y0 - R/k

So the solution is
y(t) = R/k + (y0 - R/k)e-kt

Now look at this solution for the three different scenarios. Each one will have an effect on the sign of the exponential term. You're not going to get precise graphs, since you don't know the values of R and k, but you should get a good idea of the general behavior.

1. What is exponential decay?

Exponential decay is a mathematical concept that describes the decrease of a quantity over time, where the rate of decrease is proportional to the current value of the quantity. This means that the larger the quantity, the faster it will decrease.

2. What is the formula for exponential decay?

The formula for exponential decay is N(t) = N0e-kt, where N(t) is the quantity at time t, N0 is the initial quantity, e is the base of the natural logarithm, and k is the decay rate constant.

3. How do you solve a differential equation for exponential decay?

To solve a differential equation for exponential decay, you can use separation of variables. This involves separating the variables on each side of the equation and integrating both sides. You can then solve for the constant term using initial conditions.

4. What is the half-life of a substance undergoing exponential decay?

The half-life of a substance undergoing exponential decay is the amount of time it takes for half of the initial quantity to decay. This can be calculated using the formula t1/2 = ln(2)/k, where k is the decay rate constant.

5. How is exponential decay used in real life?

Exponential decay is used in various fields such as physics, chemistry, biology, and finance. It can be used to model radioactive decay, population growth, and decay of natural resources. It is also used in financial modeling to describe the depreciation of assets over time.

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