# hw2

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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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