MATH 54 QUIZ 10 SOLUTIONS
APR 23, 2014
GSI: MINSEON SHIN
(Last edited April 24, 2014 at 11:31am.)
Problem 1. State Euler’s identity. Solve for cos x and sin x in terms of eix and e−ix .
Solution. Euler’s identity is
eix = cos x + i sin x
which implies
e−ix = cos x − i sin x
and
1 ix 1 −ix
e + e
2
2
1 ix −1 −ix
sin x = e +
e
.
2i
2i
cos x =
Problem 2. Find a linear second-order constant-coefficient homogeneous ODE satisfied by y(x) = e3x cos 2x.
Solution. We know that if the auxiliary equation to ay 00 + by 0 + cy = 0 has two complex conjugate roots α ± βi, then the
space of solutions to ay 00 + by 0 + cy = 0 is the span of eαx cos βx and eαx sin βx. Thus we need to find a polynomial at2 + bt + c
which has the root 3 + 2i. It suffices to consider the polynomial (t − (3 + 2i))(t − (3 − 2i)) = t2 − 6t + 13. The associated
linear second-order constant-coefficient homogeneous ODE is y 00 − 6y 0 + 13y = 0.
Problem 3. Find the general solution to the differential equation
y 00 + 2y 0 − 3y = xex .
(1)
Solution 1: Undetermined coefficients. We first find a particular solution to (1). Refer to [1, page 422]. The auxiliary
polynomial is x2 + 2x − 3 = (x − 1)(x + 3). In the notation of the textbook, we have r = 1, which has multiplicity s = 1 in
the auxiliary polynomial x2 + 2x − 3. Thus our trial solution is yp (x) = x(ax + b)ex . We then have
yp (x) = (ax2 + bx + 0)ex
yp0 (x) = (ax2 + (2a + b)x + b)ex
yp00 (x) = (ax2 + (4a + b)x + (2a + 2b))ex
so that
yp00 + 2yp0 − 3yp = (a + 2a − 3a)x2 + ((4a + b) + 2(2a + b) − 3(b))x + ((2a + 2b) + 2(b) − 3(0))1 ex
= ((8a)x + (2a + 4b))ex .
Comparing with (1) gives 8a = 1 and 2a + 4b = 0, which implies a = 81 and b = −1
16 . Thus our particular solution is
1 2
1
x
00
yp (x) = ( 8 x − 16 x)e . The general solution to the associated homogeneous ODE y + 2y 0 − 3y = 0 is yc (x) = c1 ex + c2 e−3x .
Thus the general solution to (1) is
y(x) = ( 81 x2 −
1
x
16 x)e
+ (c1 ex + c2 e−3x ) .
Solution 2: Annihilator method. Suppose y(x) is a solution to (1), which we can rewrite as
(D − 1)(D + 3)y = xex .
(2)
We find a differential operator which annihilates xex . In general, we have that (D − r)m+1 annihilates P (x)erx where P (x)
is a polynomial of degree m. Thus (D − 1)2 annihilates xex . Thus we have
(D − 1)3 (D + 3)y = 0 .
(3)
The general solution to (3) is yc (x) = (c0 + c1 x + c2 x2 )ex + c3 e−3x . Then
(D − 1)(D + 3)yc = (D − 1)(D + 3)((c0 + c1 x + c2 x2 )ex + c3 e−3x )
= (D + 3)(D − 1)((c0 + c1 x + c2 x2 )ex ) + (D − 1)(D + 3)(c3 e−3x )
= (D + 3)((c1 + 2c2 x)ex ) + 0
= ((D − 1) + 4)((c1 + 2c2 x)ex )
= (2c2 )ex + 4(c1 + 2c2 x)ex
= ((2c2 + 4c1 ) + 8c2 x)ex
and comparing this with (2) gives 8c2 = 1 and 2c2 + 4c1 = 0, which implies c2 =
to (2) is
y(x) = (c0 −
1
16 x
1
8
and c1 =
−1
16 .
Thus the general solution
+ 18 x2 )ex + c3 e−3x .
(Note: The coefficients c0 , c1 , c2 , c3 in this solution are not related to the c1 , c2 in Solution 1.)
Problem 4. Solve the initial value problem
y 00 − y = sin x − e2x ,
(4)
0
y(0) = 1, y (0) = −1.
Solution 1: Undetermined coefficients. We find a particular solution to y 00 −y = sin x and a particular solution to y 00 −y = e2x ,
then subtract them to obtain (by the superposition principle) a particular solution to (4). Refer again to [1, page 422].
We compute a particular solution to y 00 − y = sin x. Since i is not a root of the auxiliary polynomial x2 − 1, our trial solution
is yp (x) = a sin x+b cos x. We have yp00 (x) = −a sin x−b cos x, so yp00 −yp = −2a sin x−2b cos x. Comparing with y 00 −y = sin x
gives −2a = 1 and −2b = 0, which implies a = − 21 and b = 0. Thus a particular solution to y 00 − y = sin x is yp (x) = − 21 sin x.
We compute a particular solution to y 00 − y = e2x . Since 2 is not a root of the auxiliary polynomial x2 − 1, our trial solution
is yp (x) = ae2x . Then yp00 (x) = 4ae2x , so that yp00 − yp = 3ae2x . Comparing with y 00 − y = e2x gives 3a = 1, which implies
a = 13 and that yp (x) = 13 e2x is a particular solution to y 00 − y = e2x .
The above implies that yp (x) = − 21 sin x − 13 e2x is a particular solution to (4). The general solution to the associated
homogeneous ODE y 00 − y = 0 is c1 ex + c2 e−x , so the general solution to (4) is yc (x) = (− 12 sin x − 13 e2x ) + (c1 ex + c2 e−x ). The
condition yc (0) = 1 implies c1 + c2 = 34 . We have yc0 (x) = (− 21 cos x − 32 e2x ) + (c1 ex − c2 e−x ), and the condition yc0 (0) = −1
7
implies c1 − c2 = 16 . Solving these two equations for c1 , c2 gives c1 = 34 and c2 = 12
. Thus the solution to the IVP is
y(x) = (− 12 sin x − 13 e2x ) + ( 34 ex +
7 −x
)
12 e
.
Solution 2: Annihilator method. Suppose y(x) is a solution to (4), which we can rewrite as
1
1
(5)
(D − 1)(D + 1)y = eix − e−ix − e2x .
2i
2i
We have that D−i annihilates eix , D−(−i) annihilates e−ix , and D−2 annihilates e2x , so their product (D−i)(D−(−i))(D−2)
1 ix
1 −ix
annihilates 2i
e − 2i
e
− e2x . Thus we have
(D − i)(D − (−i))(D − 2)(D − 1)(D + 1)y .
(6)
y(x) = c0 eix + c1 e−ix + c2 e2x + c3 ex + c4 e−x .
(7)
The general solution to (6) is
It’s true in general that if p(t) is any polynomial, then p(D)(erx ) = p(r)erx .1 Then we have
(D − 1)(D + 1)y = (D2 − 1)y
= c0 ((i)2 − 1)eix + c1 ((−i)2 − 1)e−ix + c2 ((2)2 − 1)e2x + c3 ((1)2 − 1)ex + c4 ((−1)2 − 1)e−x
= −2c0 eix − 2c1 e−ix + 3c2 e2x
and comparing with (5) gives −2c0 =
1
2i ,
1
−2c1 = − 2i
, and 3c2 = −1. This means that
−1 ix
1
1
e + e−ix − e2x + c3 ex + c4 e−x
4i
4i
3
−1
1 2x
sin x − e + c3 ex + c4 e−x
=
2
3
and the rest is similar to that of Solution 1.
y(x) =
1For example, (D 2 + 2D + 3)(e5x ) = (52 + 2 · 5 + 3)e5x .
(8)
(9)
Problem 5. True or False? Justify your answer (if True, provide a proof; if False, provide a counterexample).
(i) Functions y1 , . . . , ym are linearly independent on (0, 1) if and only if they are linearly independent on (−1, 1).
(ii) Let y1 , . . . , ym be functions on R. Then W [y1 , . . . , ym ](x) = 0 for all x ∈ R if and only if y1 , . . . , ym is linearly
dependent.
(iii) Suppose y1 and y2 are solutions to the ODE y 00 + 2y 0 − 6y = 0 on R. If y1 (0) = y2 (0) and y10 (0) = y20 (0), then
y1 (x) = y2 (x) for all x ∈ R.
(iv) If y1 and y2 are solutions to y 00 + y = sin x, then c1 y1 + c2 y2 is also a solution to y 00 + y = sin x for any scalars c1 , c2 .
Solution.
(i) False. It is true that if y1 , . . . , ym are linearly independent on (0, 1), then they are linearly independent on
(−1, 1). (In other words, if y1 , . . . , ym are linearly dependent on (−1, 1), then they are linearly dependent on (0, 1).)
But the converse is false. Consider the case m = 2, and y1 (t) = t and y2 (t) = |t|. Then y1 , y2 are linearly dependent
on (0, 1) (since they are equal on (0, 1)), but y1 , y2 are linearly independent on (−1, 1). We can prove this as follows.
Suppose there exist constants c1 , c2 such that c1 t + c2 |t| = 0 for all t ∈ (−1, 1). Taking t = 21 , we get 12 c1 + 21 c2 = 0.
Taking t = − 21 , we get − 12 c1 + 12 c2 = 0. These two equations imply that c1 = c2 = 0.
(Note: I should have made this clear, but it is implicit that y1 , . . . , ym are all defined on (−1, 1).)
(ii) False. It is true that if y1 , . . . , ym is linearly dependent, then W [y1 , . . . , ym ](x) = 0 for all x ∈ R. But the converse is
false. Consider the case m = 2 with y1 (t) = t2 and y2 (t) = t · |t|. Then y1 and y2 are both differentiable2 and
2
t t · |t|
W [y1 , y2 ](t) = det
=0
(10)
2t 2|t|
for all t. But y1 , y2 are linearly independent on R for the same reason that t, |t| are linearly independent on (−1, 1),
as in (i).
(iii) True. This follows from the uniqueness part of [1, page 403, Theorem 1].
(iv) False. For example, the function y(x) = 0, obtained as the linear combination 0y1 + 0y2 , is not a solution to
y 00 + y = sin x.
Here is a related true statement: “If y1 and y2 are solutions to y 00 + y = 0, then c1 y1 + c2 y2 is also a solution to
00
y + y = 0 for any scalars c1 , c2 .” This should remind you of the case for homogeneous/nonhomogeneous systems of
linear equations.
References
[1] Lay, D. C., Nagle, R. K., Saff, E. B., Snider, A. D. Linear Algebra & Differential Equations, Second Custom Edition for University of California,
Berkeley. Pearson, 2012.
2It doesn’t matter that y is not twice-differentiable, since the Wronskian (of two functions) considers only the first derivative.
2