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Unformatted text preview: CME 102, Winter 2009 Ordinary Differential Equations for Engineers Prof. Eric Darve Homework 1 Solutions 1. Kreyszig: (a) exercise 6 page 8 This is a second order differential equation. We take twice the derivative of y to get y 00 = a 2 cos( x ) b 2 sin( x ) = 2 [ a cos( x ) + b sin( x )] = 2 y. We can thus conclude that y 00 + 2 y = 0. (b) exercise 16 page 8 Substitution of y l = cx c 2 into the ODE gives ( y l ) 2 xy l + y l = c 2 xc + ( xc c 2 ) = 0 thus proving that y l = cx c 2 is the general solution. Similarly, for y p = 1 4 x 2 , we get that ( y p ) 2 xy p + y p = 2 4 x 2 x 2 4 x + 1 4 x 2 = 1 4 1 2 + 1 4 x 2 = 0 . In figure 6, we can see that the parabola represents the function y p . Such parabola is the envelope of the set of general solutions, the lines y l (recall that c is an arbitrary constant). For a given c , the corresponding line y l is tangent to the parabola y p at exactly one point. (c) exercise 20 page 8 From the physical information, we have that the pressure satisfies an ODE of the form y = ky . The general solution to this ODE is y = y exp( kx ), where y is the pressure at sea level x = 0. Now we also know that y (18000) = 1 2 y = y exp( k. 18000)...
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This note was uploaded on 02/13/2009 for the course CME 102 taught by Professor Darve,e during the Winter '09 term at Stanford.
 Winter '09
 Darve,E

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