1. JL Sem2_2013/2014
TMS2033 Differential Equations
Mid-Semester Exam SOLUTION
Semester 2 2013/2014
31st
March 2014 9:00 – 11:00am
Answer all questions.
1. Determine whether any of the functions
1 2 3
1
( ) sin 2 , ( ) , ( ) sin 2
2
y x x y x x y x x  
is a solution to the initial-value problem
4 0; (0) 0, (0) 1y y y y     .
(8 marks)
.2sin4,2cos2 11 xyxy  Hence, )(1 xy is a solution to the differential equation.
Also, it satisfies the first initial condition, 0)0(1 y . However, )(1 xy does not satisfy
the second condition since 10)0(1 y . This means that )(1 xy is not a solution to the
initial value problem.
[3 marks]
.0,1 22  yy )(2 xy satisfy both the initial conditions but is not a solution to the
differential equation, hence, )(2 xy is not a solution to the initial value problem.
[2 marks]
.2sin2,2cos 33 xyxy  upon substitution into the differential equation we have
02sin
2
1
42sin2 





 xx , which shows that )(3 xy satisfy the differential equation.
Also, 0)0sin(
2
1
)0(3 y and 1)0(2cos)0(3 y , shows )(3 xy satisfy both initial
conditions. This means that )(3 xy is a solution to the initial value problem.
[3 marks]
2. Find the solution to the initial value problem, and specify the interval where the
solution is defined.
.100)2(;4
210
2


 QQ
tdt
dQ
(15 marks)
Solve using Method of integrating factor where 
dttp
e
)(
 .
Here, .,
210
2
)( 210
2



 
dt
t
e
t
tp 

2. JL Sem2_2013/2014
Let tu
u
du
dt
t
dtdutu 210lnln
210
2
2210 

  
te
t
210
210ln



Multiplying the differential equation by μ, we obtain
 
   tQt
dt
d
tQ
dt
dQ
t
840210
8402210


Upon integrating both sides above, we have
 
t
Ctt
tQ
CttQt
210
440
)(
440210
2
2




Applying the initial condition directly, we have
1304
961400
14001680
100
)2(210
)2(4)2(40
)2(
2







C
C
C
C
Q
Thus,
t
tt
tQ
210
1304404
)(
2



This solution is defined as long as 50210  tt
3. Consider the initial value problem
0(4 )/(1 ), (0) 0.y ty y t y y     
a. Find the solution to the initial value problem
t
yty
dt
dy



1
)4(
The differential equation is separable and hence,
  


dt
t
t
dy
yy 1)4(
1
To assist in integration, we write the integral terms in partial fractions:
4
1
)4(1
4)4(
1




BAByyA
y
B
y
A
yy
1,1)1(
11




BABtAt
t
B
A
t
t
Thus, our integration now becomes:

3. JL Sem2_2013/2014
441ln4
4
)1(
4
1ln4
4
ln
1ln444lnln
1ln4ln
4
1
ln
4
1
1
1
1
)4(4
1
4
1
4












 
tKee
y
y
Ctt
y
y
Cttyy
cttyy
dt
t
dy
yy
tCtt
Applying 0)0( yy  ,
K
y
y

 40
0
Therefore the solution for the initial value problem is
  4
0
4
0
144 ty
ey
y
y t



b. Determine how the solution behaves as t  
As t  , the right-hand-side of the solution found in part a. is also  .
This means that

 4y
y
And so,
4
04


y
y
c. If y0 = 2, find the implicit function of time T at which the solution first reaches
the value 3.99
 
 
  T
T
T
eT
T
e
T
e
44
4
4
4
4
1399
1
399
12
2
01.0
99.3






(20 marks)

4. JL Sem2_2013/2014
4. Consider the differential equation
222
3 yxyx
dx
dy
x 
a. Show that the equation is homogeneous
Writing the differential equation in ),( yxf
dx
dy
 form, we have
2
22
3
),(
x
yxyx
yxf

 .
),(
33
),( 2
22
22
22222
yxf
x
yxyx
xt
ytxytxt
tytxf 




Hence, the equation is homogeneous.
b. Solve the differential equation
Let y = xv then
dx
dv
xv
dx
dy
 . Substitute these into the differential equation
we obtain
2
2
2
2222
21
31
3
vv
dx
dv
x
vv
x
vxvxx
dx
dv
xv




This can be solved using method of separable variables:
 
Cx
v
x
dx
v
dv
x
dx
vv
dv










ln
1
1
1
21
2
2
Substitute back
x
y
v  , then we have
)(ln)(ln
))(ln(
ln
1
1
CxxxCxy
xCxxy
Cx
x
y





c. For a given initial condition as y(1) = 2, determine the solution for this initial
value problem.
Applying y(1) = 2 into the general solution obtained earlier, to find C

5. JL Sem2_2013/2014
3
1
12



C
CC
Then substitute this back into the general solution to obtain the solution to the
initial value problem
xxxy  )
3
1
)(ln( .
(18 marks)
5. Consider the differential equation
.022

dx
dy
bxexye xyxy
a. Find the value of b for which the given equation is exact.
Writing the DE in Mdx + Ndy = 0 form, we have
xyeM xy
 2
and xy
bxeN 2

For exact DE, we need
xy NM 
Thus,
1
)12()12(
)2(0)2(
22
2222



b
xybexye
beeybxNeexyM
xyxy
xyxy
x
xyxy
y
b. With the value of b found in part b., find the explicit solution to the
differential equation.
Let the general solution of the DE be cyx ),( , where
xy
y xeN 2

Integrate this wrt y,
)1()(
2
)(
2
)(
2
2
2
xh
e
xh
x
xe
dyxe
xy
xy
xy


 


Also,
)2(2
xyeM xy
x 
Differentiate (1) wrt x and equate with (2), we have

6. JL Sem2_2013/2014
2
)(
)(
2
2
2
2
2
x
h
xxh
xyexh
ye xy
xy
x



Substitute h(x) above into (1), we obtain
22
22
xe xy

Hence, the explicit solution to the differential equation is
Cxe xy
 22
(19 marks)