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(MATH152)midterm1.pdf
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Math 152 Midterm Exam
October 27, 2000
Your Name
Student Number
1.
Feel free to ask if you are confused about problems.
2.
Do not look at your book and notes. For more space, write on the opposite side.
3.
Show all your work. Cross o. (instead of erase) the undesired part.
4.
Provide all the details. Your reason counts most of the points.
Number Score
1
2
3
4
Total
1
y . + y = f(t).
where . . 0if |t|> 1
.
f(t)=
.
.
1if |t|< 1.
Solution. An integrating factor is exp[ dt]= et. Multiplying the equation by et,weget
tt)
(ye = f(t)e.
t
So ye= f(t)etdt,or
.t
y = ef(t)e tdt
The di.culty of this problem is to compute f(t)etdt. We can do it this way:
t f(t)e tdt = C + f(s)e sds.
0
tt
When |t|1, we have f(s)esds = 0 esds = et .1.
0
t 1
When t 1, we have f(s)esds = 0 esds = e .1.
0
t .1
When t .1, we have f(s)esds = esds = e.1 .1. Therefore, the general solution is
00
y = Ce.t + yp(t)
where
.
. e1.t if t 1
.
.
.
.
.
yp(t)=1 if |t|1
.
.
.
.
.
.
e.1.t if t .1.
Note that yp is a continuous function. One can solve the equation in the three regions (t 1, |t| 1 and t .1) separately and then glue the solutions in the three regions together continuously to obtain the general solution of the equation.
y . + y = sin t.
(Hint: You may need to use L{f(t) g(t)} = L{f(t)}L{g(t)}.)
Solution. We shall .nd a solution yp with yp(0) = y(0) = 0. Taking the Laplace Transform of
p
the equation, we get
s 2L(yp)+ L(yp)= 1 .
s2 +1
So
11
L(yp)= .
s2 +1 s2 +1
So
tt
11
yp = sin t . sin t = sin(t . ) sin d = [ [cos(t . 2) . cos t]d = [sin t . t cos t].
0 20 2
3 (20 points) Suppose y1 =1, y2 =1+ t and y2 =1+ t2 are three solutions of a linear second order di.erential equation.
(1)
Write down the general solution of this linear second order di.erential equation.
(2)
Find this linear second order di.erential equation.
Solution.
(1) y2 . y1 = t, and y3 . y1 = t2 are two linear indepedent solutions of the corresponding homogeneous equation, so the general solution of this linear second order di.erential equation is
y = C1t + C2t +1
where C1 and C2 are arbitrary.
(2) Let the equation be y. +py. +qy = g. Since t, t2 are solutions of equation y+py+qy =0, we have
(t). + p(t). + qt =0, 2). + p(t
(t2). + qt2 =0. 2
Then we must have p = .2 and q = . Since y1 = 1 is a solution of y. + py. + qy = g, we have
tt2
. . 2 . + 22
g = q. So, the equation is yyt2 y = t2 , or (since we allow t = 0.)
t
2
ty . . 2ty. +2y =2.
y . + iy = 2 cos t
it + e
(Here i is the imaginary unit. Note that 2 cos t = e.it.) by using
(1)
The Method of Integrating Factor.
(2)
The Method of Variation of Parameters.
(3)
The Method of Undetermined Coe.cients.
(4)
Laplace Transform.
Solution.
.it .it it(eit +e.it
(1) An integrating factor is eit. So, y = eeit2 co