This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Solutions to Assignment 2 - MAT1332A - Fall 2009 (1) (a) [6 points] Check that N ( t ) = t 1 + ct is a solution of the differential equation dN dt = N 2 t 2 . Treat c as an unspecified constant. (b) [5 points] Use N (1) =- 1 to find c . Then give the solution N ( t ) corre- sponding to this initial condition. Solution: (a) On one hand, dN dt = d dt t 1 + ct = 1 + ct- ct (1 + ct ) 2 = 1 (1 + ct ) 2 , and on the other hand, N 2 t 2 = t 2 (1 + ct ) 2 1 t 2 = 1 (1 + ct ) 2 . Hence dN dt = N 2 t 2 . (b) N (1) = 1 1 + c =- 1 = - 1- c = 1 = c =- 2 Therefore N ( t ) = t 1- 2 t . (2) The rate at which a bacteria population multiplies is proportional to the instantaneous amount of bacteria present at any time. The mathematical model for this dynamics can be formulated as follows: db dt = kb, where b is a function in terms of the time t , b ( t ) is the number of bacteria at the time t , and k is a constant. The general solution of this autonomous differential equation is b ( t ) = b (0)...
View Full Document
- Fall '09