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Unformatted text preview: 18.03 Lecture #32 Nov. 23, 2009: notes PROBLEM SET 9 is due Monday, November 30 (not 11/25 as originally announced). Problem set 10 will still be due Friday 12/4. Topic for today is systems of ordinary differential equations. So far we have considered equa tions where one dependent variable changes as a function of one independent variable. Examples include ... Independent var. Dep. var. Equation Description S = amt of material t = time S ′ + kS = 0 radioactive decay T = temperature t = time T ′ = k ( T T ) cooling y = height x = horiz. dist. ay ′′ = radicalbig 1 + ( y ′ ) 2 hanging chain (EP page 46) p = pressure h = height dp dh = ρp Air pressure vs. height x = position t = time mx ′′ + cx ′ + kx = f forced springanddashpot I = current t = time LI ′′ + RI ′ + 1 C I = E ′ RLC circuit All of this is close to the world of onevariable calculus. There are two ways to generalize this. One would be to allow several independent variables; for example to look at the strength of an electromagnetic field as a function of position in space (three variables). This leads to partial differential equations, and problems that are different and more complicated in fundamental ways. What we will do is stay in the world of one independent variable (very often time ), but allow several dependent variables . That is, we will look at several things all changing together. Here are some typical examples. Motion in several dimensions. Simplest example is the motion of a planet around the sun, discussed in EP at the end of Section 5.1. If we think of the sun (having a large mass M ) as fixed at the origin, and the planet as having mass m and position x ( t ) = ( x 1 ( t ) ,x 2 ( t ) ,x 3 ( t )) , then Newton’s law of gravity says that the gravitational force between them is F = mMG ( x 2 1 ( t ) + x 2 2 ( t ) + x 2 3 ( t )) 3 / 2 x ( t ) . (Why the 3 / 2 exponent on the distance term in the denominator? What happened to the “inverse square” law?) So Newton’s law of motion says that x ′′ 1 = mMG ( x 2 1 ( t ) + x 2 2 ( t ) + x 2 3 ( t )) 3 / 2 x 1 x ′′ 2 = mMG ( x 2 1 ( t ) + x 2 2 ( t ) + x 2 3 ( t )) 3 / 2 x 2 x ′′ 3 = mMG ( x 2 1 ( t ) + x 2 2 ( t ) + x 2 3 ( t )) 3 / 2 x 3 (Planetary motion) In order to solve them for the precise motion of a planet, we need to know the position and velocity of that planet at one time t : x 1 ( t ) = x 1 , x 2 ( t ) = x 2 , x 3 ( t ) = x 3 , x ′ 1 ( t ) = v 1 , x ′ 2 ( t ) = v 2 , x ′ 3 ( t ) = v 3 . (Initial conditions) 1 (Quite a bit of additional physical and mathematical insight is needed to put these equations in the form appearing in EP and then actually to solve them.) One bit of evidence that these equations are not fundamentally different from the ones we’ve already studied is that Euler’s method for finding approximate solutions still applies. If we divide time into little intervals of length h , then we can approximate the velocities at time t + h by linear approximations using the second derivatives that we know: for example...
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 Fall '09
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 Differential Equations

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