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Unformatted text preview: Math 2a Prac Lectures on Differential Equations Prof. Dinakar Ramakrishnan 272 Sloan, 25337 Caltech Oce Hours: Fridays 45 PM * Based on notes taken in class by Stephanie Laga, with a few added comments 1 Lecture 1 A differential equation, often shortened as diffEq or just DE, is an equa tion involving the derivatives of a function y = f ( t ), where t is an independent variable . Here the dependent variable y could be just a single variable, which is what we will consider at the beginning, or a vector ( y 1 ,...,y n ) in nspace; our main focus will be on n = 2 , 3. The diffEq is called ordinary , and written as ODE, unless the equation contains partial derivatives, in which is case it is called a partial differential equation, abbreviated as PDE. In this course we will deal mostly with ODEs, though some partial derivatives will at times be used in their analysis. The order of an ODE is the highest order of the derivatives which appear. Examples : (1) In classical Mechanics, Newtons law (in 1 dimension) says F = ma, a = dv dt , t . It is a first order ODE in v , the velocity. Moreover, v = dx dt , where x is the position, so the force is F = m d 2 x dt 2 , and we obtain a second order ODE in the independent variable x . Suppose you want to solve F = m dv dt , with fixed mass m . One can break up into three cases: (i) F = 0 (no force): dv dt = 0 = v = v , the initial value : This is the principle of inertia: If there is no force acting on a particle, it stays its course, i.e., remains at rest (for v = 0), or it moves at constant nonzero speed (when v = 0); the sign of v gives the direction. (ii) F = c = 0 (constant nonzero force) e.g. c = mg : the case of a freely falling particle F = m dv dt = mg = v = gt + v 2 v = dx dt = x = g t 2 2 + v t + x , where x is the initial position. (iii) F nonconstant: F = mg v m dv dt = mg v Two forces : mg drag = v, : constant = ( v mg ) Stationary point: v = mg dv dt = 0 To get nonstationary solutions v = mg , divide by v mg = 0 and obtain 1 ( v mg ) dv dt = m 1 v mg ( dv dt ) dt = m dt + C Want to find a function ( v ) such that d dv = 1 v mg By applying the chain rule, d dt = 1 v mg dv dt , implying ( v ) = ln ( v mg ) up to a constant. We get ln  v mg  = m t + C , which we exponentiate to get v mg = e ln( v mg ) = e mt + C = Ae m t , where A = e C > v mg = Ae m t 3 So, including the stationary solution as well, we get v = mg + Be m t , where B is any constant; B = 0 corresponds to the stationary state....
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This note was uploaded on 12/02/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.
 Fall '08
 Makarov,N
 Equations

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