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(math150)[2010](s)final~946^_10469.pdf
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Final Exam MATH 150: Introduction to Ordinary Differential Equations 27th May 2010
Answer ALL questions Full mark: 100 Time: 16:30-18:30
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:
.
This exam is for TWO HOURS.

.
Please find the Table of Laplace transforms in the last page.


Question No. (mark) Marks
1 (10)
2 (10)
3 (20)
4 (20)
5 (20)
6 (20)
Total (100)

1. Solve the initial value problem
y= 2(1 + x)(1 + y2), y(0) = 0 for y ()
= yx and then determine where the solution attains its minimum value. (Hint: Use the substitution y =tan z for integration.)
Solution:

2. Find the solution y=() yx of the initial value problem
+ 2 x
y 4y= x + 3e, y(0) = 0, y'(0) = 2 using the method of undetermined coefficients.
Solution:

3. Use the Laplace transform to find the solution =()yyt of the initial value problem:
13 t/2
y y 4 y=(t +1)e , y(0) =1, y(0) =0.
.+
Solution:
Step1: We take the L. T. of the equation
3 t/2
{}. {} +1 Ly =Lt +1 e
Ly Ly 4 {}{( )}.
Using the L.T. table, we have
2 13 t/2
() .sy(0) .y(0)] [ s() .y(0)] + () =Lt (+1)e }.
[sYs . Ys Ys {
4 Substituting the two initial conditions into the above equation, we can get an algebraic equation for ()
Ys:
..s2 .+s 1 ..Ys () .+= s 1 {(3 +)et/2 }.
Lt 1 . 4 . The Laplace transform of the inhomogeneous term is
3 t/2 3 t/2 t/26 1
Lt( +)e }=Lte {+Le = 4
{ 1 }{} + .
(s.1/ 2 ) s.1/ 2
Then,
.4 .1
Ys 6(s.1/ 2 ) +(s.1/ 2 ) s 1
+.
() = (s.1/ 2 )2 .
61 s.1
=++
632
(s.1/ 2 )(s.1/ 2 )(s.1/ 2 ) Step2: Taking the inverse L. T.
.1
yt() =L Ys { ()}
..
. 61 s.1 .
=L.1 . 6 + 3 + 2 .
.(s.1/ 2 )(s.1/ 2 )(s.1/ 2 ).

.. . .. ...
1 .1 . 5! . 1 .1 . 2! ..1 . s.1 .
L . 51++ . 21 . L .

= 20 .(s.1/ 2 ) .. 2 L .(s.1/ 2 )+.+ .(s.1/ 2 )2 ..
. .. ... .... ..
1 .1 . 5! . 1 .1 . 2! ..1 . 1 . 1 .1 . 1! .. 51+. 21 .. 11
= L .+ L + .+L .. L +.
20 ..(s.1/ 2 ) .. 2 ..(s.1/ 2 ).. .s.1/ 2 . 2 ..(s.1/ 2 ) ..
15 t/2 12 t/2 t/2 1 t/2
= te + te +e . te
202 2 We conclude that the solution of the initial value problem is
t/2 .1 5121 .
yt()=e . t + t . t+1..
.20 2 2 .
4. Given the initial value problem: y+4 = ()
y gt ,
.cos()t,0 <4 t
where gt()=..1, < t 2, y(0) =0, y(0) =1. .
0, 2<
t
.
(1)
Express the function gt() in terms of the step (Heaviside) function.

(2)
Take the Laplace transform of gt().


=
(3) Find the solution yyt() of the given initial value problem.
Solution:
(1) ()=cos4 1.ut ()+ ().u
gt t. .. ut (t)...
. .. 2
(2) The Laplace transform of gt()is
{ ()}=L{cos 4 .cos .4 t.).ut+ ()
Lgt t ( () ut.u (t)}
.. 2 { } { .( ). () } {()} {2()}
=Lcos 4 t.Lcos .4 t. .ut +Lu t .Lu t .
.s .2s
s .se .e
= 2 (1.e )+
s+16 s
(3) We first take the Laplace transform of the equation {''}+4{} Ly = { ( )} .
Ly Lgt