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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 )(yfdt
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()( 2yyyf 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.
231 yf
For max/min point, let .0)( yf We get, 5773.03
113 2 yy and
3849.033
2
3
1
f
Now, yf 6 . Then,
33
2,
3
10
3
6
3
1f is a max point
33
2,
3
10
3
6
3
1f is a min point
JL Sem2_2013/2014
d. Draw the phase line and classify the equilibrium solutions obtained in part (b).
ydt
dyyfyfor 00)(,1, increasing
ydt
dyyfyfor 00)(,01, decreasing
ydt
dyyfyfor 00)(,10, increasing
ydt
dyyfyfor 00)(,1, decreasing
Therefore, the equilibrium point at 0y is unstable and the equilibrium point at
1y are stable.
e. Sketch the direction fields of the differential equation, ).1( 2yydt
dy Make sure the
concavity of the solution curves is determined and the locations of the inflection
points are found.
JL Sem2_2013/2014
For concavity of the solution curves, we need to determine 22 dtyd . We have,
)()()()(2
2
yfyfdt
dyyfyf
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 when3
1y . Hence, for concavity:
,1y 0f and f > 0 02
2
dt
ydconcave down
3
11 y , 0f and f < 0 0
2
2
dt
ydconcave up
0,03
1 fy and f < 0 0
2
2
dt
ydconcave down
3
10 y , 0f and f > 0 0
2
2
dt
ydconcave up
0,13
1 fy and f > 0 0
2
2
dt
ydconcave down
,1y 0f and f < 0 02
2
dt
ydconcave 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
