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AMATH_383_Solution_1

# AMATH_383_Solution_1 - #1Winter2010 Problem2.4 a N c t m t...

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AMATH 383 Solution to Assignment #1 Winter 2010 Problem 2.4 a. 3/4 0 ( ) ( ) / . . ( ) . c c c c c c N t m t m E d d E N m dt m dt Y Y m Therefore the equation c c c c d Y Y N E N dt Becomes: 3/4 0 ( ) . c c c c Y E d Y m m m m m dt This is the same as: 3/4 0 ( ) ( ) c c c c Y m Y d m m dt E E . m So 0 / and / . c c c c a Y m E b Y E b. m=M when dm/dt=0. From the above differential equation we find: 3/4 0 . aM bM Solving for M : 4 ( / ) . M a b Now back to the time-dependent differential equation, we replace b/a by M -1/4 to get: 3/4 1/4 1 ( / ) . d m am m M dt Let 1/4 3/4 1/4 1/4 1/4 ( / ) , and 1 . . 4 1 . 4 ( ) . 4 r m M R r d m d r m dt M dt d a r r dt M d a R R dt M   Solve for R(t) :

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