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					                             Differential Equation Answer
1—3
                                                                    dy
4.                                                                      t 1
                                                                          2
             the slope field of equation
                                                                    dt

                                                the slope field of Dy=t 2+1
                   2


              1.5


                   1


              0.5


                   0


              -0.5


                   -1


              -1.5


                   -2
                     -2      -1.5         -1     -0.5          0         0.5       1         1.5       2




7.(a)
                          solution of Dy=3*y*(1-y) with y(0)=-1,0.5,1,2,3 on the slope field
              2


            1.5


              1


            0.5

                                                        the solution with initial conditiony(0)=0.5
              0
     y(t)




            -0.5


             -1


            -1.5


             -2
               -2          -1.5      -1        -0.5       0        0.5         1       1.5         2
                                                           t



(b)
from the figure of (a)
we can see that when the t is increasing, then the solution with initial
condition y(0)=0.5 is approaches the equilibrium value 1.
10.(a)
                             solution of Dy=(t+1)*y with y(0)=-1,0.5,1,2,3 on the slope field
                 2


            1.5


                 1


            0.5
                                                                      the initial condition y(0)=0.5
   y(t)




                 0


           -0.5


                 -1


           -1.5


                 -2
                  -2         -1.5      -1        -0.5      0         0.5      1        1.5        2
                                                            t



(b) from the figure of (a)
we can see that when the t is increasing, then the solution with initial
condition y(0)=0.5 is also increasing .
11.(a)

                      2


                  1.5


                      1


                  0.5
          y(t)




                      0


                 -0.5


                      -1


                 -1.5


                      -2
                        -2     -1.5         -1     -0.5         0      0.5        1       1.5          2
                                                                 t



                                                                                                           dy
from the above figure we can know that when t=1 ,the                                                          is 0, and
                                                                                                           dt
                                      dy                   dy
when t >1, then                          <0 .when t<1,then    >0,so we can choose the
                                      dt                   dt
                                         dy
differential equation (4)                    1 t .
                                         dt
(b)

               2


             1.5


               1


             0.5
  y(t)




               0


             -0.5


              -1


             -1.5


              -2
                -2    -1.5   -1   -0.5       0    0.5       1    1.5    2
                                              t


from the above figure we can know that when y=1 is a equilibrium
                                            dy                  dy
value ,and when y>1, then                      <0.when y<1,then    >0 . so we can
                                            dt                  dt
                                                         dy
choose the differential equation (5)                         1 y .
                                                         dt
(c)

                2


              1.5


                1


              0.5
      y(t)




                0


             -0.5


               -1


             -1.5


               -2
                 -2   -1.5   -1   -0.5       0     0.5      1     1.5   2
                                              t



from the above figure we can know that when y=1,y=-1 is a equilibrium
                                         dy                   dy
value ,and when y>1,then                    >0.when y<-1,then    >0,when -1<y<1,
                                         dt                   dt
              dy                                                   dy        2
then             <0 ,so we can choose the differential equation (8)     y        1.
              dt                                                   dt
(d)



                2


              1.5


                1


              0.5
       y(t)




                0


              -0.5


               -1


              -1.5


               -2
                 -2   -1.5   -1   -0.5    0    0.5   1   1.5   2
                                           t



this slope field changes as y and as t change, so it can be equations (3) or
(8) . when t=0 the slope are positive so it must be equation (3)

16.
  (a) by the definition the slope field we can sketch the slope field from
the solution figure with initial condition y(0)=1.
  (b) the solution figure of y(0)=2 is the solution figure with initial
condition y(0)=1 shift up to y(0)=2.because the differential equation
depends only on t.

17.
   (a) because the equation is autonomous ,the slope field is constant on
horizontal lines ,so this solution is enough information to sketch the
slope field on the entire upper half plane ,also if we assume that f is
continuous then the slope field on the line y=0 must be horizontal.
   (b) the solution with initial condition y(0)=2 is a translate to the left
of the solution given.
1—8
2.
     dy 3
        y t
              5

     dt t
     dy 3
        y t
              5

     dt t


                                              3
the integrating factor  (t )  e t dt 
                                                        1
                                                            3
                                                        t
                                                         1
product both side of equation with                           3
                                                         t

         1 dy 3
             4 y t
                      2
          3
         t dt t
                 1
      d(                y )  t dt
                                  2
                     3
                 t
                     1
       d(                  y )   t dt
                                          2
                         3
                     t
                              3

                 yt
         1
            3
                                  C
         t                   3
                         6

     yt                     Ct
                                      3

                         3
         (C is a constant)
4.

          2ty  4 e t
      dy                  2



      dt
          2ty  4 e t
      dy                2



      dt



the integrating factor  (t )  e
                                                             2

                                                       et
                                              2 tdt



                                                                            2

product both side of equation with product both side of equation with   t
                                                                        e
                2
                  dy        2

            et       2t et y  4
                  dt
                   2

            d ( t y )  4dt
                e
                          2

             d (et          y )   4dt
                 2

            et y  4t  C
                               2            2

            y  4t et  C et

               (C is constant)
12.
           dy   2t        2
                     y        , y (0)  2
           dt 1  t 2
                         1 t
                              2


           dy   2t        2
                     y
           dt 1  t 2
                         1 t
                              2




                                                               2t
the integrating factor  (t )  e 1t dt                       2
                                                                          1
                                                                         1 t
                                                                                2




product both side of equation with product both side of equation with
  1
1 t
       2




                1         dy            2t                           2
                                                    y
               1 t                 (1t 2)                (1t 2)
                      2                           2                       2
                          dt

                1                            2
            d(       y)                                 dy
               1 t                   (1t )
                    2                                 2
                                                  2

                          1                       2
             d(                  y)                        dt
                     1 t                   (1t )
                               2                           2
                                                      2



now we want to find the left side integrating
let
       tta
           n
       dt  s e cd
                         2


                     2
                s e cd
                                    2

                
                         4
           sec
         2 c o s d
                 
                              2



                                1
         (1  c o 2 )d    s i n
                    s                2
                                2


                                                                       1         2t
and we know that t  tan    tan t and sin 2 
                                                                                 1 t
                                                                                        2




                 2                            1        t
so                           dt  tan t 
           (1t )                                  1 t
                         2                                  2
                     2

hence the solution of equation is
  1                          1      t
           y  tan t                         C
1 t                               1 t
       2                                  2


                              1
y  (1  t ) tan t  t  C (1  t )
                 2                                  2




by the initial condition y(0)=-2 ==> C =-2
so the solution with initial condition y(0)=-2 is
                              1
y  (1  t ) tan t  t  2(1  t )
                 2                                 2



18.
                             dy
                                 y  4 cost
                                             2

                             dt
                             dy
                                 y  4 cost
                                             2

                             dt




      the integrating factor  (t )  e
                                                                1dt        t
                                                                       e

      product both side of equation with product both side of equation with
            t
           e
                  t   dy    t      t
                           e y  4 e cost
                                           2
              e        dt
                             t                            t
               d (e y )  4 e cost dt
                                                                         2


                                  t                            t
                d (e y )   4 e cost dt
                                                                             2


                        t                            t
               e y   4 e cost dt
                                                                     2


                                                  t
               y  4e                     e
                                       t                         2
                                                       cost dt


 22.

               dS
                    0.06S  10000
               dt
               S (0)  30000
               dS
                    0.06S  1 0 0 0 0
               dt


      the integrating factor  (t )  e
                                                                                      0.06tdt        0.06t
                                                                                                  e

product both side of equation with product both side of equation with
    0.06 t
e

                        0.06 t   dS           0.06 t            0.06 t
                   e                  0.06S e          10000 e
                                  st
                                    0.06 t                                       0.06 t
                    d (e                        S )  10000 e                             dt
                                            0.06 t                                    0.06 t
                     d (e                           S )    10000 e                          dt
                             10000 0.06t
                              0.06 t
                   e                     C
                                           S
                              0.06 e
                       10000
                   S        Ce
                                   0.06 t

                        0.06

by the initial condition S(0)=30000
                                              10000                   10000
                       30000                        C  C  30000                                            1.3667 10
                                                                                                                          5
                                                                            =
                                               0.06                    0.06
                                  10000
                       S                (1.3667 10 )  e
                                                      5     0.06t

                                   0.06
now we want to find out the t when S(t)=0
S (t )  0  (1.6667 10 )  (1.3667 10 )  e                              0
                                           6                    5   0.06 t



    1.6667 10                                     1.6667 10
                      6                                                 6
                                             1
                         e           t 
                              0.06 t
                                                ln
    1.3667 10                              0.06 1.3667 105
                      5


t  3.3075( years)
24.
      dG          1
          2          G
      dt        15  t
      G(0)  6
      dG      1
                 G2
      dt 15  t
                                                            1
    the integrating factor  (t )  e 15t  15  t
                                           dt




    product both side of equation with product both side of equation

 with 15+t

             dG
      (15  t )    G  2(15  t )
              dt
       d ((15  t )G )  (30  2t )dt
        d ((15  t )G )   (30  2t )dt

       (15  t )G  30t  t  C
                                               2


                                       2

                 t 
           30t            C
      G
          15  t 15  t 15  t

    by the initial condition G(0)=6
       C
 6        C  90
       15
                          2

           t 
     30t           90
 G
    15  t 15  t 15  t


                       30t  90
                  2

      G(t )  t                                    and when t=15 it will full
                   15  t
                  15  450  90
                     2

       G(15)                                       25.5
                          15  15
so 25.5 pound in the tank when it is full.
   1—6
   4.
                                             the plot of f(w)=w*cos(w)
                       4


                       3


                       2


                       1
                f(w)




                       0


                       -1


                       -2

                                              the equilibrium value
                       -3


                       -4
                         -5   -4   -3   -2      -1      0       1        2   3   4   5
                                                        w



   from the plot of f(w)=w*cos(w) , we know that when
                n
   w=0,w=          ,n   ,f(w)=0
                 2
   from the figure
   w=0, is a source
         
   w= ,           , is a sink
        2   2
        3  3
   w=      ,    , is a node
         2   2


22.(a)
      we can discuss six situation:
(1) when y=2 is node, y=-1 is sink:
    the solution of y(t) with initial condition y(0)=1 is decreasing from
    equilibrium value 2 to equilibrium value –1 as t decreasing.
(2) when y=2 is node, y=-1 is source:
    the solution of y(t) with initial condition y(0)=1 is increasing from
    equilibrium value -1 to equilibrium value 2 as t increasing
(3) when y=-1 is node, y=2 is sink:
     the solution of y(t) with initial condition y(0)=1 is like situation (2).
(4) when y=-1 is node, y=2 is source:
  the solution of y(t) with initial condition y(0)=1 is like situation (1).
(5) when y=-1 is sink, y=2 is source:
  the solution of y(t) with initial condition y(0)=1 is like situation (1).
(6) when y=-1 is source, y=2 is sink:
  the solution of y(t) with initial condition y(0)=1 is like situation (2).

(b)
      like situation (a)(2).

 24.
   from the figure there are three equilibrium value denoted A,B,C
   A>B>C
   A is node ; B is source ; C is sink.
 26.
   from the figure there are four equilibrium value denoted A,B,C,D
   A>B>C>D
   A is sink ; B is node ; C is source ; D is sink.
 36.(a)
                                         the plot of 1/(y 2-y-2)
                   20



                   10



                     0
            f(y)




                   -10



                   -20



                   -30



                   -40
                      -2   -1.5    -1   -0.5       0         0.5   1   1.5   2
                                                   y



      dy                                   dy
                                  y  1 ,
                                        
y2 ,                                       
      dt                                   dt
    dy                                  dy
y2 ,                          y  1 ,     
      dt                                   dt
                                            dy        1
 (b) solve the equation                        
                                            dt ( y  2)( y  1)
 ( y  2)( y  1)dy  dt
        2
 ( y  y  2)dy  dt
        3         2


    y       
              y        2y  t  C
    3         2
     1
y ( 0) 
     2
      13
C 
      12
        3         2


    y       
              y        2y  t 
                                      13
    3         2                       12
plot the figure of the solution

                                3

                              2.5

                                2

                              1.5

                                1
                       y(t)




                              0.5

                                0

                              -0.5

                               -1

                              -1.5

                               -2
                                2.5   2    1.5   1   0.5   0      -0.5   -1   -1.5   -2   -2.5
                                                           t

				
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