This preview shows page 1. Sign up to view the full content.
Unformatted text preview: , such as time t ; and which are dependent , such as y = y ( t ). #2. Articulate the principle that underlies the problem under investigation. This can be used to articulate the diFerential equation. #3. Identify the initial conditions. An initial value problem consists of (1) a diFerential equation and (2) a list of initial conditions. 1.2: Solutions of Some Dierential Equations. We review how to solve an initial value problem in the form dy dt = b + a y DiFerential Equation y (0) = y Initial Condition for some constants a and b : dy dt = a y b a 1 y b a dy dt = a d dt ln y b a = a = ln y b a = a t + C 1 for some constant C 1 . Upon exponentiating both sides we nd that y ( t ) b/a = C 2 e at for some constant C 2 = e C 1 . or the initial condition, we set t = 0: C 2 = y b a = y ( t ) = b a + y b a e at . 1...
View
Full
Document
This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue UniversityWest Lafayette.
 Spring '09
 EdrayGoins
 Basic Math

Click to edit the document details