Math 286 Spring 2016
Practice Final Exam Solutions
(1) Solve the initial value problem, where primes denote derivatives with
respect to x,
(
y 0 = y sin x,
y(0) = 1.
Solution Separating variables and integrating both sides, we have
Z
Z
1
dy = sin xdx,
y
ln |y| = − cos x + C,
y = e− cos x · D,
for some constants C and D. Since y(0) = 1, it follows that D = e.
Hence the solution is
y(x) = e1−cos x .
(2) Find the general solution of the equation
3
y 0 + 3x2 y = e−x sin x.
Solution This is a first-order linear differential equation with integrating factor
R
2
3
ρ(x) = e 3x dx = ex .
Multiplying both sides by ρ(x) yields
3
3
(ex y)0 = ex (y 0 + 3x2 y) = sin x.
Then we integrate both sides to obtain
Z
x3
e y = sin xdx = − cos x + C,
3
y = e−x (C − cos x),
where C is a constant.
(3) Suppose that the characteristic equation of a differential equation is
(r2 + 1)(r − 2)2 = 0.
(a) Find the differential equation associated to this characteristic
equation.
(b) Find the general solution of the differential equation obtained
from (a).
Solution
(a) Since (r2 +1)(r−2)2 = (r2 +1)(r2 −4r+4) = r4 −4r3 +5r2 −4r+4,
the associated differential equation is
y (4) − 4y (3) + 5y 00 − 4y 0 + 4y = 0.
(b) The roots of the characteristic equation are r = ±i, 2, 2. Hence
the general solution is
y(x) = A cos x + B sin x + Ce2x + Dxe2x .
(4) Consider an undamped system consisting of a mass of 4 kg on a
spring with constant k = 4 N/m, and external force F (t) = 2 sin ωt.
(a) Let ω = 2. Set up the nonhomogeneous equation which describes
this system and find a particular solution.
(b) Find the external frequency ω which causes resonance.
Solution
(a) The nonhomogeneous differential equation describing this system
is
4x00 + 4x = 2 sin 2t → 2x00 + 2x = sin 2t,
where x(t) is the position function. Since the characteristic equation
of the associated homogeneous equation has roots r = ±i, we try
xp = A cos 2t + B sin 2t.
Plugging into the equation, we have
2x00p + 2xp = −8A cos 2t − 8B sin 2t + 2A cos 2t + 2B sin 2t
!
= −6A cos 2t − 6B sin 2t = sin 2t.
Hence A = 0 and B = −1/6. Therefore, we have a particular solution
1
xp = − sin 2t.
6 p
√
(b) Resonance occurs when ω = k/m = 1 = 1.
(5) Let


1 1 1
A = 1 1 1 .
1 1 1
(a) Find all eigenvalues of A and their defects.
(b) Find the general solution of the system
~x0 (t) = A~x(t).
Solution
(a) The characteristic polynomial is
1 − λ
1
1 1−λ
1 = −λ3 + 3λ2 = −λ2 (λ − 3).
|A − λI| = 1
1
1
1 − λ
Thus the eigenvalues are λ = 0, 0, 3.
λ = 0 We find the eigenvectors by solving

   
1 1 1 a
0
(A − 0 · I)~v = 1 1 1  b  = 0 .
1 1 1
c
0
Hence we have only one linear equation, namely, a + b + c = 0, or
c = −a − b. It follows that we can write ~v as


 
 
a
1
0
~v =  b  = a  0  + b  1  ,
−a − b
−1
−1
so ~v1 = [1 0 − 1]| and ~v2 = [0 1 − 1]| are linearly independent
eigenvectors associated to λ = 0.
λ = 3 We find the eigenvectors by solving

   
−2 1
1
a
0





b = 0 .
(A − 3 · I)~v = 1 −2 1
1
1 −2
c
0
Applying row operations,

1 −2
0 1
0 0
one sees that this system is equivalent to
   
1
a
0
−1  b  = 0 .
0
c
0
Hence b − c = 0 and a − 2b + c = 0, so b = c and a = 2b − c = c.
Therefore, we have
 
 
c
1



~v = c = c 1 ,
c
1
so ~v3 = [1 1 1]| is the eigenvector associated to λ = 3.
Since the number of eigenvectors equals the multiplicity of λ, the
eigenvalues have defect 0.
(b) Since the eigenvalues are complete, we have the general solution
~x(t) = c1 v~1 + c2 v~2 + c3 v~3 e3t .
(6) Find a particular solution of the differential system
(
x0 = 4x + y − 1,
y 0 = x + 4y − et .
Solution We first rewrite the system as
4 1
−1
0
0
~x =
~x +
+
,
1 4
0
−et
4 1
where ~x = [x
The characteristic equation of A =
is
1 4
|A − λI| = λ2 − 8λ + 15 = (λ − 3)(λ − 5), so the eigenvalues are
λ = 3, 5. Hence we take a particular solution of the form
a
b
t
~xp = ~a + ~be = 1 + 1 et .
a2
b2
y]| .
Plugging ~xp into the equation yields
t
t
~be = A(~a + ~be ) + −1 + 0 et .
0
−1
By comparing coefficients, we have
4 1 a1
1
3 1 b1
0
~
A~a =
=
,
(A − I)b =
=
.
1 4 a2
0
1 3 b2
1
Upon solving these two systems, we find that a1 = 4/15, a2 =
−1/15, b1 = −1/8, b2 = 3/8.
(7) (a) Suppose that f is a function of period 2 with f (t) = t for 0 <
t < 2. Show that
∞
2 X sin(nπt)
f (t) = 1 −
π
n
n=1
(b) Substitute an appropriate value of t to deduce
π
1 1 1
= 1 − + − + ···
4
3 5 7
Solution.
(a) We compute the Fourier coefficients of f
Z
a0 =
2
t dt = 2
0
2
cos(2nπ) + 2nπ sin(2nπ) − 1
=0
n2 π 2
0
Z 2
sin(2nπ) − 2nπ cos(2nπ)
2
bn =
t sin(nπt) dt =
=−
2
2
n π
nπ
0
Z
an =
t cos(nπt) dt =
Since f is a piecewise smooth function we know that its Fourier series
converges, and is equal to f (t) for 0 < t < 2. Thus
f (t) = 1 −
∞
2 X sin(nπt)
.
π
n
n=1
(b) Substituting t = 1/2 shows that
∞
∞
2 X sin(nπ 21 )
2 X sin(nπ 12 )
=1−
π
n
π
n
n=1
n=1
2
1 1 1
=1−
1 − + − + ···
π
3 5 7
f (1/2) = 1/2 = 1 −
hence
π
1 1 1
= 1 − + − + ···
4
3 5 7
(8) Find the steady periodic solution xsp (t) of the equation x00 (t) +
5x(t) = F (t), where F (t) is the function of period 2π such that
F (t) = 3 if 0 < t < π and F (t) = −3 if π < t < 2π.
Solution. The Fourier series of F (t) has coefficients
Z
1 2π
a0 (F ) =
F (t) dt = 0
π 0
Z
1 2π
F (t) cos(nt) dt = 0
an (F ) =
π 0
Z
Z
Z
1 2π
3 π
3 2π
bn (F ) =
F (t) sin(nt) dt =
sin(nt) dt −
sin(nt) dt
π 0
π 0
π π
6
= (1 − (−1)n )
π
sin(nt)
12 P
and hence F (t) = π
n odd
n . We substitute this into the differential
equation and also expand xsp (t) into its sine series xsp (t) =
P∞
b
(x
n
sp ) sin(nt) to find
n=1
−
∞
X
n2 bn (xsp ) sin(nt) + 5
n=1
∞
X
bn (xsp ) sin(nt) =
n=1
12 X sin(nt)
.
π
n
n odd
Thus we have bn (xsp ) = 0 for n even, and bn (xsp ) = 12/πn(5 − n2 )
for n odd, showing that xsp (t) is given by
xsp (t) =
12 X sin(nt)
.
π
n(5 − n2 )
n odd
(9) The edge x = a of the rectangular plate 0 < x < a, 0 < y < b, is
insulated, the edges x = 0 and y = 0 are held at temperature zero,
and u(x, b) = f (x). Use the odd half-multiple sine series to derive a
solution of the form
nπx nπy X
u(x, y) =
cn sin
sinh
2a
2a
n odd
Solution. The boundary value problem is


uxx + uyy = 0, for 0 < x < a, 0 < y < b
u(0, y) = ux (a, y) = u(x, 0) = 0


u(x, b) = f (x)
The eigenvalue problem
(
X 00 + λX = 0,
X(0) = X 0 (a) = 0
yields
λ = λn =
(2n − 1)2 π 2
,
4a2
Xn (x) = sin
(2n − 1)π
x ,
2a
n = 1, 2, . . .
Then
Yn00 − λn Yn = 0
(2n − 1)π
(2n − 1)π
=⇒ Yn (y) = An cosh
y + Bn sinh
y
2a
2a
and Yn (0) = 0 implies An = 0. Thus we have as general solution
∞
X
(2n − 1)π
(2n − 1)π
u(x, y) =
x sinh
y .
Bn sin
2a
2a
n=1
We find the coefficients Bn by imposing u(x, b) = f (x), which requires
b
2n−1
Bn =
sinh (2n−1)π
b
2a
where {b2n−1 } are the coefficients of the odd half-multiple sine series
of f (x) on [0, a].
(10) Consider the Sturm-Liouville problem
(
y 00 (x) + λy(x) = 0, for 0 < x < L
y(0) = 0, hy(L) − y 0 (L) = 0
where h and L are positive constants satisfying hL = 1.
Show that the eigenvalues are λ0 = 0 and λn = αn2 for n = 1, 2, . . .
where αn are the positive solutions of tan(αL) = αh . What are the
associated eigenfunctions?
Solution. First consider λ < 0, say λ = −α2 with α > 0.
The general solution to y 00 (x) + λy(x) = 0 is y(x) = A cosh(αx) +
B sinh(αx) and the boundary conditions impose
y(0) = A = 0,
Since h =
1
L
hy(L) − y 0 (L) = hB sinh(αL) − αB cosh(αL) = 0.
we can rewrite the second equation as
B
cosh(αL) (tanh(αL) − αL) = 0
L
and since the only solution of tanh(t) = t is t = 0, we conclude that
B = 0. Hence λ < 0 is not an eigenvalue.
Next consider λ = 0. The general solution to y 00 (x) = 0 is A + Bx
and the boundary conditions impose A = 0 and h(A + BL) − B =
hA + (hL − 1)B = 0. Thus we see that λ0 = 0 is an eigenvalue with
eigenfunction y0 (x) = x.
Finally consider λ > 0, say λ = α2 with α > 0. The general
solution to y 00 (x) + λy(x) = 0 is y(x) = A cos(αx) + B sin(αx) and
the boundary conditions impose
y(0) = A = 0,
hy(L) − y 0 (L) = hB sin(αL) − αB cos(αL) = 0.
If B 6= 0 then the second equation can be rewritten as
α
tan(αL) =
h
and so the positive eigenvalues are λn = αn2 where αn are the positive solutions of tan(αL) = α/h. The associate eigenfunctions are
sin(αn x).