now use to find
Then we can discuss b)
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 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.