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Score:15/15
Math 54 Lecture 1-Quiz 10, August 7th, 2013
Any answer with no justification will get zero points. Each problem worth 5 points.
1. Determine whether the set of functions
{sin x, cos x, tan x}
is linear independent on (− π2 , π2 ).
Solution: The Wronskian of the three functions is

sin x

W = W [sin x, cos x, tan x] = det 
 cos x
− sin x
which equals to



sin x
cos x − sin x
1
−
det 
tan x det 
2x
cos
− sin x
− sin x − cos x
= − tan x −
cos x
tan x
− sin x
1
cos2 x
2 sin x
cos3 x
− cos x
fact, when 0 < x <




sin x
2
sin
x
+
det 
3x
cos
− cos x
cos x
cos x

cos x
− sin x
2 sin x
= − tan x(1 + 2 sec2 x).
cos3 x
This is not identically zero on (− π2 , π2 ). For example, at x =
π
2,

W is always negative and when
− π2
π
4,
W (x) = −1 − 4 = 5. In
< x < 0, W is always positive.
Therefore this set of functions is linear independent.
(Alternatively, we already know that sin x, cos x are linear independent since the two functions has Wronskian −1 or they are not multiple to each other. Any linear combination of them will be a bounded function as both of them are bounded. However, tan x
is not bounded on (− π2 , π2 ), therefore tan x is not a linear combination of sin x, cos x.
This gives the linear independence of the three functions. You could also choose several
points in (− π2 , π2 ) to get a system of equations and show there is only trivial solution for
c1 sin x + c2 cos x + c3 tan x = 0. However, you have to do it correctly. ± π2 are not in the
interval and you cannot choose them.)
2. Find the general solution to
(D + 2)3 (D2 + 4D + 5)2 [y] = 0.
1


Here D =
d
dt .
Solution: The auxiliary equation is
(r + 2)3 (r2 + 4r + 5)2 = 0.
So r = −2 is a triple root and both r = −2 ± i are double roots. Therefore the general
solution is
y = C1 e−2t + C2 te−2t + C3 t2 e−2t + C4 e−2t cos t + C5 e−2t sin t + C6 te−2t cos t + C7 te−2t sin t.
3. Find the general solution to
y 00 − 2y 0 + y = t−1 et , t > 0.
Solution: Since the general solution to the corresponding homogeneous equation y 00 − 2y 0 +
y = 0 is given by
y = C1 et + C2 tet .
We look for solution in the form
y = u1 et + u2 tet .
By the method of variation of parameters,
u01 et + u02 tet = 0, u01 et + u02 et (t + 1) = t−1 et .
We can solve for u01 , u02 to get
u01 = −1, u02 = t−1 .
Therefore
u1 = −t + C1 , u2 = log t + C2 .
We have the general solution
y = −tet + tet log t + C1 et + C2 tet .
(Or we can replace C2 by C2 + 1 to get a somewhat simpler expression
y = tet log t + C1 et + C2 tet .
Alternatively, we can directly use the substitution y = uet to reduce the problem to u00 = t−1
and integrate twice to get the solution.)
2