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(MATH150)[2010](s)midterm~1806^_10470.pdf
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Midterm Exam MATH 150: Introduction to Ordinary Differential Equations 27th March 2010
Answer ALL questions Full mark: 100; each question carries 20 marks Time allowed C 1 hour
This is a closed book exam. You may write on the front and back of the exam papers.
Student Name:
Student ID Number:
Lecture section (L1 or L2):

Question No. (mark) Marks
1 (20)
2 (20)
3 (20)
4 (20)
5 (20)
Total

1. Find the solution =()
uur of the initial value problem:
du 1+u2
= 3, u(1) =1, r >0
.
dr ru +ru
Solution:
udu rdr
=
22 2
1+ur (1+r )
u du2 r dr2
=
2 22
11+u 1 r (1+r )
u .1+u2 .
... =ln1 21 ..
LHS (+u )=ln
2
..
22 2
rdr rdr 2 .1+r .
... =. =ln ()
RHS r .ln .
12 1 r22 .
r 1+
..
4r2 u2 =.1 1+r2 Using the conditionu(1) =1, we have
4r2
u =
.1 1+r2.
2. Find the solution =()yyt of the initial value problem: dyt (t 1) y =t, y(ln 2) =1, t >0.
++
dt
Solution:

3. A culture (i.e., a collection of organisms) has N0 bacteria at time t = 0 initially. At t = 1 hour, the number of bacteria is measured to be 3 N0 . If the
2 rate of growth of the bacteria number is proportional to the number of bacteria present, determine the time necessary for the number of bacteria to triple.
Solution:
The number of bacteria is governed by the equation dN / dt = kN with N(0) = N0.
The solution is Nt() = N e kt , in which the rate constant k is determined through the

0
relation N(1) = N e k = 3N /2 : k = ln(3 / 2) . The time for N to triple is determined by
00
k ln3 ln3
e = 3 . It follows that == .2.7 hours.
k ln(3 / 2)
4. Find the solutions x=xt() of the following initial value problems:
(a) x.6x+9x=0, x. =. 1, x.=1
( 1) (1) .
(b) x.6x +10 x=0, x(0) =0, x(0) =1.
Solutions:
(a) Using the Ansatz solution x() t =et, we get the characteristic equation
2 .6+9 =0,
from which we have 1 =2 =3.
The general solution is

t
x()t =(c+ct e .
12 ) 3
The first derivative of the general solution is
t
) 3
x()t =(3c+c +3ct e .
12 2
Using the given two conditions, we have
xt( =.1) = c1 .ce .3 =.1,
( 2 )
xt=.1) =( c+c .3 ).3
(3 ce =1.
12 2
33
ec 4. So the solution for the initial value problem is given by Then, we get c1 =3, 2 = e
)
3(t 1)
xt() =(3 +4te + .
(b) Use the Ansatz solution x() t =et, we get the characteristic equation
2 .6+10 =0, from which we have 3
=i.
1,2
The general solution is x()t =e3t(Asin t+Bcos t) and its first derivative is
x()t =e3t(3Asin t+3Bcos t+Acos tB . sin t).
Using the given two conditions, we have A=1, B=0. So the solution for the initial value problem is given by x()t =e3tsin t.
5. Use the method of undetermined coefficients to find the general solution yyt() of the equation
=
2
dy dy 2t
. 4 + 4y = 6te .
dt2 dt
Solution: