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midterm_1_review (dragged)

# midterm_1_review (dragged) - such as time t and which are...

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MA 36600 MIDTERM #1 REVIEW Chapter 1 § 1.1: Some Basic Mathematical Models; Direction Fields. Newton’s Second Law of Motion is the statement “ F = m a ;” it really means m d 2 x dt 2 = the sum of the forces on the object.” Newton’s Law of Gravitational Attraction is the statement “any body with mass M attracts any other body with mass m directly toward the mass M , with a magnitude proportional to the (product of the two) masses and inversely pro- portional to the square of the distance separating them.” Another way to say this is F M m r 2 = F = G M m r 2 for some (universal) constant G . If M and r are the mass and radius of the Earth, respectively, then F = m g in terms of g = G M/r 2 = 9 . 81 m / sec 2 = 32 . 5 ft / sec 2 . In general say that we have an equation in the form dy dx = G ( x, y ) . A direction field or a slope field is a plot in the xy -plane where for each point ( x, y ) we plot an arrow with slope G ( x, y ). There are three key steps to using a di ff erential equation to model a physical situation: #1. Identify the key variables. Decide which variables are
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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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