1
Spring 2019
Assignment Section 3.2 The Wronskian
2
(1 point) Determine whether the following pairs of functions
are linearly independent or not on the whole real line.
? 1. f(t)= t and g(t)= |t|
? 2. f(t)= t2 +18t and g(t)= t2 −18t
? 3. f(θ)= 18cos3θ and g(θ)= 72cos3 θ−54cosθ
Answer(s) submitted:
•Linearly independent
•Linearly independent
•Linearly dependent
(correct)
Correct Answers:
•LINEARLY INDEPENDENT
•LINEARLY INDEPENDENT
•LINEARLY DEPENDENT
2. (1 point) Use the Wronskian to show that the
functions y1 = e6x and y2 = e4x are linearly independent.
Wronskian = det
These functions are linearly independent because the
Wronskian is [Choose/zero/nonzero] for all x.
Answer(s) submitted:
•exp(6x)
•-2exp(10x)
•nonzero
(correct)
Correct Answers:
•<table border=’0’ cellspacing=’10’>
<tr><td> eˆ(6*x) </td><td> eˆ(4*x) </td></tr>
<tr><td> 6*eˆ(6*x) </td><td> 4*eˆ(4*x) </td></tr>
</table>
•eˆ(6*x)*4*eˆ(4*x)-eˆ(4*x)*6*eˆ(6*x)
•nonzero
3. (1 point) Use the Wronskian to determine whether
the functions y1 = sin(2x) and y2 = cos(5x) are linearly
independent.
Wronskian = det
These functions are linearly independent because the Wronskian is nonzero for [Choose/some/all] value(s) of x.
Answer(s) submitted:
•sin(2x)
3
•sin(2x)(-5sin(5x))-(2cos(2x))cos(5x)
•some
(correct) Correct Answers:
6. (1 point)
d2y
dx2
2 dy
3x cos(8x) = 0 on . dx
(a) What does the Wronskian of y1,y2 equal on ?
W(y1,y2) = on .
equation? [Choose/Yes/No]
Solution: (a)
By the fundamental trigonometric identity, we have that y2 = 1.
The wronksian of the two functions is then
W
(b)
Since the Wronskian equals zero, the two functions are linearly
dependent and they do not form a fundamental set for the
given differential equation. (note that linear depence can also
be shown from the fact that the two functions are scalar
multiples).
Answer(s) submitted:
•3(0)-0(1)
•No
(correct)
0
=
e+e.
1
=
3c
-
3c
with
solution
c;
=
2
and
c)
=
—2.
Thus
y=
zea
_
ze
v =
$
sinh
(32)
_
cosh
(31)
sinh(31)|
sy
easy
W(t)
=
5
sinh (37)
cosh
(31)
cosh
(0)
sinh
(1)
=1
As
expected,
the
Wronskian
is
non
zero
for
all
values
of
t,
thus
y,
and
y2
are
linearly
independent
and
they
form
a
fundamental
set
of
solutions
for
the
differential
equation.
Hence
the
general
solution
can
be
written
in
the
form
y=kiyi(t)
+koy2(t)
This
fundamental
set
is
useful
because
it
allows
to
express
the
general
solution
of
the
homogeneous
differential
equation
in
terms
of
the
initial
conditions
in
a
simple,
explicit
manner
.
More
specifically,
the
solution
that
satisfies
the
initial
condition
y(0)
=
yo
and
y'(0)
=
yo
is
given
by
y
=
yocosh
(37)
+
4yp
sinh
(32)
Answer(s)
submitted:
(1/2)
exp
(5t/4)
+
(1/2)
exp
(-5t/4)
(2/5)
exp
(5t/4)
+
(-2/5)
exp
(-5t/4)
[
((1/2)
exp
(5t/4)
+
(1/2)
exp
(-5t/4)
) (
(1/2)
exp
(St
/4)
+
(
(correct)
Correct
Answers:
e@
(1/2
-
0)
*exp((0
-
5/4)*t)
+
(1/2
+
0)
*exp((0
+
e
(0
-
2/5)
*exp((0
-
5/4)*t)
+
(0
+
2/5)
*exp((0
+
e
exp(-0/16
*t)
5.
(1
point)
Determine
the
largest
interval
in
which
the
given
initial
value
problem
is
certain
to
have
a
unique
twice-
differentiable
solution.
Do
not
attempt
to
find
the
solution.
ax
dx
sin(t)
Wm
+cos(t)—
.75)
=4,
x/(0.
ai
x(0.75)
=4,
x"(0.75
+sin(t)x=tan(t),
Interval:
Solution:
We
first
put
the
differential
equation
in
standard
form:
ax
cos(t)
dx
xe
tan(r)
dt?"
sin(t)
dt
sin(t)”
The
functions
a
and
on
are
defined
for
t
A
}
+km
and
t
#kn
with
k
=0,+1,42...
Since
the
initial
condition
is
given
at
x
=
0.75 with
0
<
0.75
<
,
the
largest
interval
in
which
the
given
initial
value
problem
is
certain
to
have
a
unique
twice
differentiable
solution
is
(03)
2
Answer(s)
submitted:
©
(0,pi/2)
(correct)
Correct
Answers:
e
(0,pi/2)
It
can
be
shown
that
y;
=
3
and
y
=
cos?(8x)
+
sin?(8x)
are
solutions
to
the
differential
equation
8x°
sin(3x)—
—
(0.8)
nu
(0%
)
(0.5)
(b)
Is
{y1,y2}
a
fundamental
set
for
the
given
differential
0
2
1
(1,92)
= i
0 =
L/2)
exp
(—5t/4))
}-[
(
(2/5)
exp
(St /4)
-
(2/5)
exp
(-5t/4)
) (
(5/8)
exp
(5t/4)
-(!
5/4)
*t)
5/4)
*t)
=2
(b)
Is
{y1,y2}
a
fundamental
set
for
the
given
differential
2
—1x
e
(1,92)
=
|
e
Tel
~
e
Yes
4Correct Answers:
5
•3*[8*2*sin(8*x)*cos(8*x)-8*2*cos(8*x)*sin(8*x)]
•No
7. (1 point)
It can be shown that y1 = e2x and y2 = e−7x are solutions to
d2y dy
the differential equation 2 +5dx −14y = 0 on (−∞,∞). dx
(a) What does the Wronskian of y1,y2 equal on (−∞,∞)?
W(y1,y2) = on (−∞,∞).
equation? [Choose/Yes/No]
Solution: (a)
The wronksian of the two functions is
x e −5x
W 2 x e
(b)
Since the Wronskian does not equal zero, the two functions are
linearly independent and they form a fundamental set for the
given differential equation.
Answer(s) submitted:
•-9exp(-5x)
•Yes
(correct)
Correct Answers:
•-9*eˆ(-5*x)
