DIFFERENTIAL EQUATION ANSWER
4—1
17. find the general solution of the equation:
d
2
dt
y
2
2
dy
t
 ye
dt
the general solution is: y 
h
p
y of the homogeneous equation : d
first we find
suppose the
y y
h
2
dt
mt
y (t ) = e
y
2
2
dy
y0
dt
, and it is satisfies the homogeneous equation
h
2
hence the characteristic equation is: m  2m  1  0 , so the characteristic
equation have the repeat root : m=-1;then we know that
t
y e
h
,and
we must to find another solution :
by the method of undetermined coefficients:
y
suppose
2
d y
t
 ue
h
2
h
dt
u
d e
dt
t
2
d
and
y
h
dt
du t
t
ue ,
e
dt

du t
t
 ue
e
dt
then we can get the equation : d
2
u
dt
y
so
t
h
y
dt
t
 (a  bt ) e  a e  bt e
second we find
d
t
 0 so u(t)=a+bt; a,b is constant
y
,guessing
p
y
p
 ct
2
t
e
,and
2
p
d y
 2ct e  c t e ,
dt
t
2
t
2
p
t
t
 2c e  4ct e  c t
2
t
e
fixed in the equation
t
t
 2c e  4ct e  c t
t
t
 2c e  e  c 
2
t
t
 4ct e  2c t
e
2
t
e
 ct
2
t
e
t
e
1
2
y
so the particular solution
of the equation is : y 
p
y y
h

1 2 t
;hence the general solution
2t e
t
p
 (a  bt ) e 
1 2 t
, a,b is constant
2t e
34.
y (t )
is the solution of equation:
1
d
2
y
dt
p
2
y (t )
is the solution of equation:
2
d
2
dt
y
2
p
2
y
dt
let y (t ) 
dy
 qy  h(t )
dt
y (t )  y (t )
show the
d
dy
 qy  g (t )
dt
1
p
is the solution of equation:
2
dy
 qy  g (t )  h(t )
dt
y (t )  y (t )
1
and
2
d ( y (t )  y (t ))  p d ( y (t )  y (t ))  q( (t ) 
y
y
dt
dt
d y (t )  p d y (t )  q (t ))  ( d y (t )  p d y (t )  q (t ))
(
y
y
dt
dt
dt
dt
d
2
2
y
dy
p
 qy 
dt
dt
1
2
1
2
1
2
1
2
1
1
2
2
 g (t )  h(t )
y (t )  y (t )
so
d
1
2
y
dt
2
p
is the solution of equation :
dy
 qy  g (t )  h(t )
dt
40.
(a) find the general solution of equation
d
2
dt
2
2
y
2
 4y  6  t  e
2
t
2
2
2
(t ))
first suppose
y e
mt
h
,so the characteristic equation is :
m
2
40
then the characteristic equation have complex root 2i ,-2i
y (t )  c cos(t )  c sin( t )
so the
1
h
2
2
second suppose
y
 (a  bt  c t )  d e ,and
2
p
t
d y
dt
2
p
 2ct  d e
t
fixed in the equation
 2c  d e  4a  4bt  4c t  4d e  6  t  e
t
2
11

a  8
2c  4a  6 
4b  0
b  0



1 
4c  1
c  4
5d  1

d  1
5

t
y
p
2
t
11 1 2 1 t
(t )  (  t )  e
8 4
5
so the general solution is
y (t ) 
y
h
(t ) 
y
p
11 1 2 1 t
(t )  c1 cos(t )  c2 sin( t )  (  t )  e
8 4
5
(b) by the initial condition y(0)  y ' (0)  0
1
1 t
y ' (t )   c1 sin( t )  c2 cos(t )  t  e
2
5
1
y ' (0)  0  c2   0
5
11 1
y (0)  0  c1    0
8 5
63

c1   40

  1
c2
5
so the solution with initial condition is:
y(t )  
63
1
11 1 2 1 t
cos(t )  sin( t )  (  t )  e
40
5
8 4
5
c after the long-time the solution of (b) is tend to
