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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
2y 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
02sin2
142sin2
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(;4210
2
tdt
dQ
(15 marks)
Solve using Method of integrating factor where dttp
e)(
.
Here, .,210
2)( 210
2
dt
tet
tp
JL Sem2_2013/2014
Let tuu
dudt
tdtdutu 210lnln
210
22210
tet
210210ln
Multiplying the differential equation by μ, we obtain
tQtdt
d
tQdt
dQt
840210
8402210
Upon integrating both sides above, we have
t
CtttQ
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
CQ
Thus,
t
tttQ
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
tdy
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
BABtAtt
BA
t
t
Thus, our integration now becomes:
JL Sem2_2013/2014
441ln4
4
)1(4
1ln44
ln
1ln444lnln
1ln4ln4
1ln
4
1
1
11
)4(4
1
4
1
4
tKeey
y
Ctty
y
Cttyy
cttyy
dtt
dyyy
tCtt
Applying 0)0( yy ,
Ky
y
40
0
Therefore the solution for the initial value problem is
4
0
4
0
144 ty
ey
y
yt
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
1399
12
2
01.0
99.3
(20 marks)
JL Sem2_2013/2014
4. Consider the differential equation
222 3 yxyxdx
dyx
a. Show that the equation is homogeneous
Writing the differential equation in ),( yxfdx
dy form, we have
2
22 3),(
x
yxyxyxf
.
),(33
),(2
22
22
22222
yxfx
yxyx
xt
ytxytxttytxf
Hence, the equation is homogeneous.
b. Solve the differential equation
Let y = xv then dx
dvxv
dx
dy . Substitute these into the differential equation
we obtain
2
2
2
2222
21
313
vvdx
dvx
vvx
vxvxx
dx
dvxv
This can be solved using method of separable variables:
Cxv
x
dx
v
dv
x
dx
vv
dv
ln1
1
1
21
2
2
Substitute back x
yv , 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
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
dybxexye 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 xybxeN 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
xhe
xhx
xe
dyxe
xy
xy
xy
Also,
)2(2 xyeM xy
x
Differentiate (1) wrt x and equate with (2), we have
JL Sem2_2013/2014
2
)(
)(2
2
2
22
xh
xxh
xyexhye 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)
