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(MATH150)finalsoln_06fall&05fall.pdf
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Please notice that, the following solutions for .nal 2005 and 2006 are only for reference and may not be totally consistent with standard grading scheme. If you .nd any questions or mistakes, please feel free to contact me.
Tutor: Liu Dongwen Email: ldxab@ust.hk
Reference solutions of final 2005
1. (a) On separating variables, we obtain
(3 + 2y)dy = 2cos2xdx. Integrate on both sides, then we have
. y . x
(3 + 2y)dy = 2cos2xdx,
.20
3y + y 2.
y = sin 2x|0 x ,
.2 y 2 +3y +2 = sin2x.
If we solve above equation for y and make use of y(0) = .2, then
.3 . 1 + 4sin2x 11
y = , 0 x . arcsin .
2 224
(b) From the result of (a), it is clear that y attains its minimum at sin 2x = 1,
or at x = . And the minimum value of y is
4
.3 . 5
ymin = . 2
2
2. (a) With
. t
t
(t) = exp dt = e,
0
we have
.. t . aa
.t
x(t)= ee t(a + be.t)dt= . e .t + bte.t .
0
(b) We set xB(t)= ae .t + be.t . bte.t =0,
then
a + b
tm = .
b It is easy to see that Bx(t) > 0 when t<tm, and Bx(t) < 0 when t>tm, thus x(t) attains its maximum at tm.
(c) Clearly, x(t) a/ as t . 2
3. Denote x(t) the temperature of the co.ee. Then with some constant k,
one has xB= k(x . 20),x(0) = 90,x(5) = 60.
Separating variables, it follows
dx
= kdt,
x . 20
. x dx . t
= dt,
90 x . 20 0
ln(x . 20)|x = kt,
90
kt
x(t) = 20+70e. Applying x(5) = 60, we have
14
k = ln
57 and
.t/5
.4
x(t) = 20+70 . 2
7
4. The characteristic equation
r 2 +2r +1=0 has a repeated root r = .1. Thus the general solution is
.t
x(t)=(c1 + c2t)e, and we have
.t
xB(t)=(.c1 + c2 . c2t)e. Applying the initial conditions gives us x(0) = c1 =1,xB(0) = .c1 + c2 = . Then c1 =1,c2 = +1 and the initial value problem has the solution
.t
x(t) = [1+( + 1)t]e.
(b) Equivalently, we need to determine such that 1+(+1)t becomes negative for large t> 0. It is obvious that < .1 is the answer. 2
5. As in the previous question, the general solution is
.t
x(t)=(c1 + c2t)e, and
.t
xB(t)=(.c1 + c2 . c2t)e.
2
The initial conditions gives that
x(0) = c1 =0,xB(0) = .c1 + c2 = u0,
or that c1 =0,c2 = u0.
Thus the solution of the initial value problem is
.t
x(t)= u0te.
(b) Now we have
.t
xB(t)= u0(1 . t)e.
Since u0 > 0, we have xB(t) > 0 when t< 1,and Bx(t) < 0 when t> 1. Thus at t = 1 the maximum value of x occurs. 2
6. First note that the general solution for x+ x =0 is
xh = A cos t + B sin t. Now we treat each time interval separately.
(i) On 0 <t<, a particular solution is xp = F0t. Then
x = xh + xp = A cos t + B sin t + F0t, xB= .A sin t + B cos t + F0. Applying the initial conditions gives that x(0) = A =0,xB(0) = B + F0 =0, or that A =0,B = .F0. Thus
x(t)= .F0 sin t + F0t.
(ii) On <t< 2, a particular solution is xp = F0(2 . t). Then x = xh + xp = A cos t + B sin t + F0(2 . t), xB= .A sin t + B cos t . F0.