Page 72 - Schaum's Outline of Differential Equations
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CHAP. 7]          APPLICATIONS  OF FIRST-ORDER  DIFFERENTIAL  EQUATIONS                55



         7.3.  What constant interest rate is required if  an initial  deposit placed into an account  that  accrues  interest
               compounded continuously  is to double its value  in six  years?
                  The balance  N(t)  in the account  at any time t is governed  by  (7.1)



               which has as its solution

               We  are  not  given  an  amount  for  the  initial  deposit,  so  we  denote  it  as  N 0.  At  t = 0,  N(0)  = N 0,  which  when
               substituted into (1) yields


               and  (1)  becomes


               We  seek  the  value  of  k  for  which  N=2N 0  when  t = 6.  Substituting  these  values  into  (2)  and  solving  for  k,
               we  find










               An interest rate of  11.55 percent  is required.


         7.4.  A bacteria culture is known  to grow  at a rate proportional to the amount  present. After  one hour,  1000
               strands  of the bacteria are observed  in the culture; and after four hours,  3000 strands. Find (a) an  expres-
               sion for the approximate number of strands  of the bacteria present in the culture at any time t and (b) the
               approximate number  of  strands  of the bacteria originally in the  culture.
               (a)  Let  N(t)  denote  the  number  of  bacteria  strands in  the  culture at time  t. From  (6.1),  dNIdt  — kN = 0, which  is
                   both linear and separable.  Its solution is



                   AU=1,N=  1000;  hence,


                   At  t = 4, W = 3000;  hence,



                   Solving (2) and  (3) for k and c, we  find




                   Substituting these values of k and c into (1), we obtain



                   as an expression  for the amount of the bacteria  present  at any time t.
               (b)  We require N at t = 0.  Substituting  t = 0 into (4), we obtain N(0)  = 694e (0 366)(0)  =  694.
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