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 Di±erential 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 ² 1 y − b a dy dt = a d dt ln ³ ³ ³ ³ y − b a ³ ³ ³ ³ = a = ⇒ ln ³ ³ ³ ³ y − b a ³ ³ ³ ³ = 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 . ²or the initial condition, we set t = 0: C 2 = y − b a = ⇒ y ( t ) = b a + ± y − b a ² e at . 1...
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 Spring '09
 EdrayGoins
 Basic Math, Mass, Constant of integration, Boundary value problem

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