Thread: Find the number of years it takes for half of the population to have been exposed to

An infectious disease spreads through a large population according to the model
dy/dt = (1-y)/4

a) Solve this differential equation, assuming y(0)=0
b) Find the number of years it takes for half of the population to have been exposed to the disease.

a) I got
y = (-1/4)Ce^t + 1

How do I find b)? Thanks

I get

now use to find

Then we can discuss b)

Originally Posted by ewkimchi

An infectious disease spreads through a large population according to the model
dy/dt = (1-y)/4

a) Solve this differential equation, assuming y(0)=0
b) Find the number of years it takes for half of the population to have been exposed to the disease.

a) I got
y = (-1/4)Ce^t + 1

How do I find b)? Thanks

What have you tried? Where are you stuck?

Step 1: .

C = 1

So I plugged it back into the equation and got

y = 1 - Ce^(t/4)

Do I make y = 1/2 and solve for t?

Originally Posted by ewkimchi

C = 1

So I plugged it back into the equation and got

y = 1 - Ce^(t/4)

Do I make y = 1/2 and solve for t?

Yes.

I have the same question for my homework but mine says that y is the percent, so y would be 50. Everything else about your question is the same so I'd guess the y value unit is too.
Also, this is the same as I got when I tried to solve it, y = 1 - Ce^(t/4), where C=1, but I'm running into a big problem. If you try to insert a y value you start by subtracting 1 from both sides.
So let's say y=50 (50 percent), 50=1-1e^(t/4)--> 49=-e^(t/4).
Divide each side by -1, -49=e^(t/4).
To solve for t, you must get rid of the e which is where I run into problems. To get rid of the e, take the natural log of each side to cancel the e. Ln(-49)=t/4 --> Ln(-49)/4=t... BUT you can't take the natural log of a negative number.

I noticed that no one has addressed this problem yet as I too have to find when half of the population is infected.

I got 1-e^-t/4

I'm sorry. I was just told that on our homework sheet the question wording was wrong. It should be y equals the portion of the population, not the percent. So y= .50, not 50

Originally Posted by NickA28

I have the same question for my homework but mine says that y is the percent, so y would be 50. Everything else about your question is the same so I'd guess the y value unit is too.
Also, this is the same as I got when I tried to solve it, y = 1 - Ce^(t/4), where C=1, but I'm running into a big problem. If you try to insert a y value you start by subtracting 1 from both sides.
So let's say y=50 (50 percent), 50=1-1e^(t/4)--> 49=-e^(t/4).
Divide each side by -1, -49=e^(t/4).
To solve for t, you must get rid of the e which is where I run into problems. To get rid of the e, take the natural log of each side to cancel the e. Ln(-49)=t/4 --> Ln(-49)/4=t... BUT you can't take the natural log of a negative number.

I noticed that no one has addressed this problem yet as I too have to find when half of the population is infected.

y is the proportion so for 50% you should put y=0.5