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Unformatted text preview: MAT 1332, Calculus for the life Sciences II Assignment 4 Solutions Problem 1: Solve the differential equation dP dt = e 2 t P 2 with initial condition P (0) = 1. Solution: First we find the general solution, then we solve for the constant using the initial condition. To find the general solution, we separate the P ’s from the t ’s, that is, multiply or divide to get all the terms with P on the left and all the terms with t on the right. That gives P- 2 dP = e 2 t dt, and so, by integrating, P- 2 dP = e 2 t dt (Note that neither side really makes sense until we integrate, since dt isn’t really a number.) Evaluating the integrals gives- 1 P = 1 2 e 2 t + C, C any constant . Note that we don’t need to include a constant for the left-hand integral (we pushed it the right-hand side of the equation). Solving for P is straightforward: P =- 2 e 2 t + 2 C C any constant . (We could also write D instead of 2 C in this equation, with D any constant.) Lastly we use the initial condition to solve for C . When t is 0, we are told that P = 1; in addition the formula says that 1 = P =- 2 e 2 · + 2 C =- 2 1 + 2 C ....
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This note was uploaded on 03/19/2011 for the course MAT 1332 taught by Professor Munteanu during the Spring '07 term at University of Ottawa.
- Spring '07