Math2065Fall-11StudentSolManual

# Math2065Fall-11StudentSolManual - William A. Adkins, Mark...

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Unformatted text preview: William A. Adkins, Mark G. Davidson ORDINARY DIFFERENTIAL EQUATIONS Student Solution Manual August 16, 2011 Springer Chapter 1 Solutions Section 1.1 1. The rate of change in the population u1D443 ( u1D461 ) is the derivative u1D443 ( u1D461 ) . The Malthusian Growth Law states that the rate of change in the population is proportional to u1D443 ( u1D461 ) . Thus u1D443 ( u1D461 ) = u1D458u1D443 ( u1D461 ) , where u1D458 is the proportionality constant. Without reference to the u1D461 variable, the differential equation becomes u1D443 = u1D458u1D443 3. Torricellis law states that the change in height, ( u1D461 ) is proportional to the square root of the height, uni221A.alt01 ( u1D461 ) . Thus ( u1D461 ) = u1D706 uni221A.alt01 ( u1D461 ) , where u1D706 is the proportionality constant. 5. The highest order derivative is u1D466 so the order is 2 . The standard form is u1D466 = u1D461 3 /u1D466 . 7. The highest order derivative is u1D466 so the order is 2 . The standard form is u1D466 = (3 u1D466 + u1D461u1D466 ) /u1D461 2 . 9. The highest order derivative is u1D466 (4) so the order is 4 . Solving for u1D466 (4) gives the standard form: u1D466 (4) = 3 uni221A.alt01 (1 ( u1D466 ) 4 ) /u1D461 . 11. The highest order derivative is u1D466 so the order is 3 . Solving for u1D466 gives the standard form: u1D466 = 2 u1D466 3 u1D466 + u1D466 . 13. The following table summarizes the needed calculations: 3 4 1 Solutions Function u1D461u1D466 ( u1D461 ) u1D466 ( u1D461 ) u1D466 1 ( u1D461 ) = 0 u1D461u1D466 1 ( u1D461 ) = 0 u1D466 1 ( u1D461 ) = 0 u1D466 2 ( u1D461 ) = 3 u1D461 u1D461u1D466 2 ( u1D461 ) = 3 u1D461 u1D466 2 ( u1D461 ) = 3 u1D461 u1D466 3 ( u1D461 ) = 5 u1D461 u1D461u1D466 3 ( u1D461 ) = 5 u1D461 u1D466 3 ( u1D461 ) = 5 u1D461 u1D466 4 ( u1D461 ) = u1D461 3 u1D461u1D466 4 ( u1D461 ) = 3 u1D461 3 u1D466 4 ( u1D461 ) = u1D461 3 To be a solution, the entries in the second and third columns need to be the same. Thus u1D466 1 , u1D466 2 , and u1D466 3 are solutions. 15. The following table summarizes the needed calculations: Function u1D466 ( u1D461 ) 2 u1D466 ( u1D461 )( u1D466 ( u1D461 ) 1) u1D466 1 ( u1D461 ) = 0 u1D466 1 ( u1D461 ) = 0 2 u1D466 1 ( u1D461 )( u1D466 1 ( u1D461 ) 1) = 2 ( 1) = 0 u1D466 2 ( u1D461 ) = 1 u1D466 2 ( u1D461 ) = 0 2 u1D466 2 ( u1D461 )( u1D466 2 ( u1D461 ) 1) = 2 1 0 = 0 u1D466 3 ( u1D461 ) = 2 u1D466 3 ( u1D461 ) = 0 2 u1D466 3 ( u1D461 )( u1D466 3 ( u1D461 ) 1) = 2 2 1 = 4 u1D466 4 ( u1D461 ) = 1 1 u1D452 2 u1D461 u1D466 4 ( u1D461 ) = 2 u1D452 2 u1D461 (1 u1D452 2 u1D461 ) 2 2 u1D466 4 ( u1D461 )( u1D466 4 ( u1D461 ) 1) = 2 1 1 u1D452 2 u1D461 uni0028.alt02 1 1 u1D452 2 u1D461 1 uni0029.alt02 = 2 1 1 u1D452 2 u1D461 u1D452 2 u1D461 1 u1D452 2 u1D461 = 2 u1D452...
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## This document was uploaded on 12/28/2011.

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Math2065Fall-11StudentSolManual - William A. Adkins, Mark...

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