# Diff EQs

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```					Diff EQs

6.6
Common Problems:

   Exponential Growth and Decay
   Compound Interest
   Newton’s law of cooling
   Other fun topics
   Suppose we are interested in a quantity that
increases at a rate proportional to the amount
present. If we also know the amount at a
certain time, we can find y as a function of
time by solving the differential equation:
dy
 ky
dx
dy
 ky
dx             y  ekteC

dy
 kdx         y  Ce   kt

y
When t = o, y =
dy
 y   kdx                     y0.
Solving for C,
C = y0

ln y  kt  C     y  y0 e kt

y  ektC         Initial Amount
The rate of change of a population of rabbits is proportional to the number of
rabbits present at any given time. If 10 rabbits are present initially, and 195
rabbits are present in 6 months, how many rabbits will there be in 2 years?

dr
 kr               r (0)  10       r (24)  ?
dt
r (6)  195
r  Ce   kt

r  10ekt
r  10e.4951t
195  10e      k ( 6)
r (24 )  10 e.4951( 24)
19.5  e  k ( 6)
r (24)  1,446,974
ln 19.5  6k
k  .4951
Suppose you deposit \$800 in an account that pays 6.3% annual interest. How
much will you have 8 years later if interest is:
A. compounded continuously
B. compounded quarterly
kt
 r
dA
 .063 A               A  800e  .063(8)        A  P1  
dt                                                       k
dA                         A  \$1324.26
 .063dt                                                             ( 4 )(8)
A                                                      .063 
A  8001    
dA                                                         4 
 A   .063dt
ln A  .063t  C                                      A  \$1319.07
A  e.063t C
A  Ce.063t
The rate at which the population of a group of organisms grows is directly
proportional to the number of organisms in the population. If the population at
time zero is 3,500 and the population after one year is 5,250, what will the
population be after 3 years?

dy                                              3
 ky                     y  3500 e
t ln
2
dt
3
y  y0 e   kt
y (3)  3500 e
( 3) ln
2

5250  3500ek (1)
y(3)  11,812.5
3
ln  k
2
Half-Life          After t years,
half of the
original amount
y  y0 e kt            is left
1
y0  y0 e kt
2
1
 e kt
2
1
ln  kt
2
ln 1  ln 2
k
t
t is the half-life.
 ln 2
k           K does not
t              depend on the
initial amount!
Example
½ life of carbon-14 is 5700 years.
Find the age of a sample in which 10% of the original radioactivity
is depleted.
 ln 2  ln 2
90% is                                          k       
still                                               t    5700
 ln 2
present                                     t
.9 y 0  y 0 e   5700

 ln 2
t
.9  e   5700

 ln 2
ln .9         t
5700

866  t
Newton’s Law of Cooling

 k T  Ts 
dT
dt

Rate of change
is   Proportional to   Difference between temp
of object’s temp
of surroundings and temp
wrt time
of object
When a murder is committed, the body, originally at 37C, cools according to
Newton’s Law of Cooling. Suppose on the day of a very difficult calc quiz, Mr.
Hopkins’ body is found at 2pm with a body temperature of 35C in a room with
a constant temperature of 20C. 2 hours later the body temperature is 30C.
Being the bright students you are, you should be able to tell Mr. Neuhard when
the heinous crime was committed.

 k T  Ts 
dT
T  20  15e kt          37  20  15e( .203)t
dt
30  20  15ek ( 2)
dT                                                       17  15e( .203)t
 T  Ts    kdt                10  15e k ( 2)
17
2                      ln  .203t
ln  2k
ln T  Ts   kt  C                  3
15

T  Ts  Ce kt                    k  .203                     .62  t

T  20  Ce                   T  20  15e( .203)t              min
 .62  60      37 min
kt

hr
35  20  Ce k ( 0)
The crime was
15  C                           committed at 1:23 pm

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