1. JL Sem2_2013/2014
TMS2033 Differential Equations
Quiz1 (2.5%)
Semester 2 2013/2014
Monday 17th
March 2014
All the best 
1. Consider the differential equation )(yf
dt
dy

a. What is the type of the differential equation above called?
Since f is a function of the unknown variable only, then this differential equation is
said to be autonomous.
b. Given ),1()( 2
yyyf  find the equilibrium solutions for the differential equation.
For equilibrium solution, dy/dt = 0. Hence, we need to solve for y in .0)1( 2
 yy
Therefore, .1010 2
 yyory
c. With the f(y) as in part (b), sketch the graph of f(y) versus y. Make sure maximum and
minimum points are clearly determined.
2
31 yf 
For max/min point, let .0)(  yf We get, 5773.0
3
1
13 2
 yy and
3849.0
33
2
3
1






f
Now, yf 6 . Then, 












33
2
,
3
1
0
3
6
3
1
f is a max point













33
2
,
3
1
0
3
6
3
1
f is a min point

2. JL Sem2_2013/2014
d. Draw the phase line and classify the equilibrium solutions obtained in part (b).
y
dt
dy
yfyfor  00)(,1, increasing
y
dt
dy
yfyfor  00)(,01, decreasing
y
dt
dy
yfyfor  00)(,10, increasing
y
dt
dy
yfyfor  00)(,1, decreasing
Therefore, the equilibrium point at 0y is unstable and the equilibrium point at
1y are stable.
e. Sketch the direction fields of the differential equation, ).1( 2
yy
dt
dy
 Make sure the
concavity of the solution curves is determined and the locations of the inflection
points are found.

3. JL Sem2_2013/2014
For concavity of the solution curves, we need to determine 22
dtyd . We have,
)()()()(2
2
yfyf
dt
dy
yfyf
dt
d
dt
dy
dt
d
dt
yd

Inflection points occur when 22
dtyd = 0. This means when f = 0. From part (c), we
obtained that f = 0 when
3
1
y . Hence, for concavity:
,1y 0f and f > 0  02
2
dt
yd
concave down
3
1
1  y , 0f and f < 0  02
2
dt
yd
concave up
0,0
3
1
 fy and f < 0  02
2
dt
yd
concave down
3
1
0  y , 0f and f > 0  02
2
dt
yd
concave up
0,1
3
1
 fy and f > 0  02
2
dt
yd
concave down
,1y 0f and f < 0  02
2
dt
yd
concave up
f. Sketch the graph of the solution to the IVP
2
1
)0(),1( 2
 yyy
dt
dy
and find the
lim ( )
t
y t

From the graph, 1)(lim 

ty
t