ok, so I dont understand how to do this. (diffeq related)
the time rate of change of a rabbit population P, is proportional to the square root of P. At t = 0 months, there are 100 rabbits, and is increasing at the rate of 20 per month. how many rabbits will there be one year later....
so how do I set this up?
1. I have dp/dt = K * sqr root(P),
2. seperate, and get dp /(sqr root(P)) = kt
3. integrate both sides
4. right side is kt + c. is the left side ln(square root p)?
5. so I get ln root(p) = kt + c.
6. get rid of ln and I have root p = De^(kt)
7. square both sides and get p = (De^(kt))^2, which is De^(2kt) right?
8. plug in t(0) = 100, I get 100 = D.
9. now to solve for k, I put t(1) = 120, and get 120 = 100e^(2k).....solve for k and k =1/2 ln (120/100), k = .0911.
10. I solve for t(12) and the equation is P = 100e^(2*.0911*12), and the population is 891. The book says the answer is 484.
Now am I totally off my rocker, or what?? anybdy?
the time rate of change of a rabbit population P, is proportional to the square root of P. At t = 0 months, there are 100 rabbits, and is increasing at the rate of 20 per month. how many rabbits will there be one year later....
so how do I set this up?
1. I have dp/dt = K * sqr root(P),
2. seperate, and get dp /(sqr root(P)) = kt
3. integrate both sides
4. right side is kt + c. is the left side ln(square root p)?
5. so I get ln root(p) = kt + c.
6. get rid of ln and I have root p = De^(kt)
7. square both sides and get p = (De^(kt))^2, which is De^(2kt) right?
8. plug in t(0) = 100, I get 100 = D.
9. now to solve for k, I put t(1) = 120, and get 120 = 100e^(2k).....solve for k and k =1/2 ln (120/100), k = .0911.
10. I solve for t(12) and the equation is P = 100e^(2*.0911*12), and the population is 891. The book says the answer is 484.
Now am I totally off my rocker, or what?? anybdy?
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