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(math150)[2004](f)final~PPSpider^sol_10463.pdf
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HKUST MATH150 Introduction to Di.erential Equations
15th Dec 2004 Final Examination Solution (Version White)
Part I: Each correct answer in the answer box forthefollowing8 multiplechoicequestions is
worth4point. DONOTguesswildly!Ifyoudonothavecon.dence inyouranswerleavethe
answer box blank. Each incorrectly answered question will result in a 0.5 point deduction.

Question 12345678 Total Answer bc cbe cad

1. Which of thefollowingfunctionscanbeused asanintegratingfactortoturnthefollowing non-exact equation into an exact equation?
dy
(3ycosx .xy sinx)+2xcosx = 0?
dx
Solution Multiplying x2y to the equation, we have an exact equation
dy

223 3
(3xy cosx .xy 2 sinx)+2xycosx =0,
dx
since
22323 3
. (3xy cosx .xy 2 sinx)=6xycosx .2xysinx = . (2xycosx).
.y.x
The answeris(b).
222 2
(a) x(b) xy (c) y(d) xy(e) xy
2. For a simple RLC series electrical circuit with R =1/5 ohm, L = 1 henry, and C farads, the di.erential equation for the current I through the circuit is
d2I 1dI
C + C + I =0 .
dt2 5 dt Pickthe largest possibleC from thefollowingwith whichthe current ofthe circuit willkeep changing its direction as t . Solution The current will oscillating between positive and negative values if the characteristic
C
equation Cr2 + r +1 =0hastwocomplexroots,i.e., C2 .4C< 0, and hence C< 100.
5 25
The answeris(c).
(a) 260 (b) 100 (c) 80 (d) 62 (e) 15
3. On which of the given intervals will the following initial value problem have a unique, continuous, solution?
dx1 2tt
ex1 tantx1(3) 2
dt t.2
=+ , = ,dx2 tet tx2 t + t2 x2(3) .3
dt
Solution Allfunctionsappearinginthesystemarecontinuousand well-de.ned everywhereexcept t (unde.ned when t = 2) and tant (unde.ned when t = +(2k +1), k =0,1,2, ).
t.22
Hence the largest interval containing the initial point t = 3 where every function in the system is
3
well-de.ned and continuous is 2 <t< 2 , on which a unique, continuous, solution is guaranteed
by the existence and uniqueness theorem.
The answeris(c).

3 3
(a) .2 <t< 2 (b) <t< 3 (c) 2<t< (d) <t< (e) 2<t< 5
2 222
4. Determine the inverse Laplace transform of the function 1
(s+2)2 . 1
Solution L{te.t}= Theansweris(b).
(s+1)2 .
.t
(a) t (b) te.t (c) t2 (d) t2 e(e) tet
5. Consider the nonhomogeneous equation
.3t
y .. +6y . +9y =(2t + t4)e.
By the method of undetermined coe.cients, there is a solution to the equation which is of the form
y = Q(t)e.3t where Q(t)is a polynomial. The degree of Q(t)is :
Solution The characteristic equation is r2 +6r +9 =(r +3)2 = 0, with repeated root r = .3.
The degree of Q(t)is 6. Or, by directly putting Qe.3t into the equation, we have

.. +6yQ.. .3t .3t
y . +9y =(Q.. .6Q. +9Q)e .3t +6(Q. .3Q)e .3t +9Qe.3t = e =(2t + t4)e,
the degree of Q must be 6.
The answeris(e).

(a)2 (b)3 (c)4 (d)5 (e)6
6. The following intial value problem of a .rst order linear system
x=3x .2y, x(0)=1,
y= .3x +4y, y(0)= .2
can be