Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 3.2 The Wronskian due 03/18/2021 at 11:59pm MST
1. (1 point) Determine whether the following pairs of func-
tions are linearly independent or not on the whole real line.
? 1. f(θ) = 10cos3θand g(θ) = 40cos3θ−30cosθ
? 2. f(t) = tand g(t) = |t|
? 3. f(t) = t2+10tand g(t) = t2−10t
Answer(s) submitted:
•Linearly dependent
•Linearly independent
•Linearly independent
(correct)
Correct Answers:
•LINEARLY DEPENDENT
•LINEARLY INDEPENDENT
•LINEARLY INDEPENDENT
2. (1 point) Use the Wronskian to show that the functions
y1=e6xand y2=e2xare linearly independent.
Wronskian = det =
These functions are linearly independent because the Wron-
skian is [Choose/zero/nonzero] for all x.
Answer(s) submitted:
•eˆ(x)
•eˆ(x)
•nonzero
(score 0.1)
Correct Answers:
•<table border=’0’ cellspacing=’10’>
<tr><td> eˆ(6*x) </td><td> eˆ(2*x) </td></tr>
<tr><td> 6*eˆ(6*x) </td><td> 2*eˆ(2*x) </td></tr>
</table>
•eˆ(6*x)*2*eˆ(2*x)-eˆ(2*x)*6*eˆ(6*x)
•nonzero
3. (1 point) Use the Wronskian to determine whether the
functions y1=sin(4x)and y2=cos(3x)are linearly indepen-
dent.
Wronskian = det =
These functions are linearly independent because the Wron-
skian is nonzero for [Choose/some/all] value(s) of x.
Answer(s) submitted:
•
•
•some
(score 0.1)
Correct Answers:
•<table border=’0’ cellspacing=’10’>
<tr><td> sin(4*x) </td><td> cos(3*x) </td></tr>
<tr><td> 4*cos(4*x) </td><td> -3*sin(3*x) </td></tr>
</table>
•sin(4*x)*(-3)*sin(3*x)-cos(3*x)*4*cos(4*x)
•some
4. (1 point)
Find the function y1of twhich is the solution of
9y00 −36y=0
with initial conditions y1(0) = 1,y0
1(0) = 0.
y1=
Find the function y2of twhich is the solution of
9y00 −36y=0
with initial conditions y2(0) = 0,y0
2(0) = 1.
y2=
Find the Wronskian
W(t) = W(y1,y2).
(Hint : write y1and y2in terms of hyperbolic sine and cosine
and use properties of the hyperbolic functions).
W(t) =
Remark: You should find that W is not zero and so y1and
y2form a fundamental set of solutions of
9y00 −36y=0.
Solution: The characteristic equation is
9r2−36 =0
with roots r1=2 and r2=−2.
The general solution is
y=c1e2t+c2e−2t,
for which
y0=2c1e2t−2c2e−2t.
The initial conditions y(0) = 1,y0(0) = 0 yield the system
1=c1+c2
0=2c1−2c2
with solution c1=1
2and c2=1
2. Thus
y1=1
2e2t+1
2e−2t=cosh(2t)
The initial conditions y(0) = 0,y0(0) = 1 yield the system
1
0=c1+c2
1=2c1−2c2
with solution c1=1
4and c2=−1
4. Thus
y2=1
4e2t−1
4e−2t=1
2sinh(2t)
W(t) =
cosh(2t)1
2sinh(2t)
2sinh(2t)cosh(2t)
=cosh2(2t)−sinh2(2t) = 1
As expected, the Wronskian is non zero for all values of t, thus
y1and y2are 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=k1y1(t) + k2y2(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) = y0and y0(0) = y0
0is given by
y=y0cosh(2t) + 1
2y0
0sinh(2t)
Answer(s) submitted:
•(1-(1/2))eˆ(2t)+(1/2)eˆ(-2t)
•(1/4)eˆ(2t)-(1/4)eˆ(-2t)
•eˆ(1+2t)
(score 0.666667)
Correct Answers:
•(1/2 - 0) *exp((0 - 2)*t) + (1/2 + 0) *exp((0 + 2)*t)
•(0 - 1/4) *exp((0 - 2)*t) + (0 + 1/4) *exp((0 + 2)*t)
•exp(-0/9 *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.
sin(t)d2x
dt2+cos(t)dx
dt +sin(t)x=tan(t),x(1.5) = 9,x0(1.5) = 6
Interval:
Solution: We first put the differential equation in standard
form:
d2x
dt2+cos(t)
sin(t)
dx
dt +x=tan(t)
sin(t).
The functions cos(t)
sin(t)and tan(t)
sin(t)are defined for t6=π
2+kπand
t6=kπwith k=0,±1,±2...
Since the initial condition is given at x=1.5 with 0 <1.5<π
2,
the largest interval in which the given initial value problem is
certain to have a unique twice differentiable solution is
0,
π
2
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•(0,pi/2)
6. (1 point)
It can be shown that y1=3 and y2=cos2(8x) + sin2(8x)
are solutions to the differential equation 8x5sin(3x)d2y
dx2−
3x2cos(8x)dy
dx =0 on 0,π
6.
(a) What does the Wronskian of y1,y2equal on 0,π
6?
W(y1,y2)= on 0,π
6.
(b) Is {y1,y2}a fundamental set for the given differential
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(y1,y2) =
2 1
0 0
=0
(b)
Since the Wronskian equals zero, the two functions are lin-
early 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:
•
•No
(incorrect)
Correct Answers:
•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=e3xand y2=e−6xare solutions to
the differential equation d2y
dx2+3dy
dx −18y=0 on (−∞,∞).
(a) What does the Wronskian of y1,y2equal on (−∞,∞)?
W(y1,y2)= on (−∞,∞).
(b) Is {y1,y2}a fundamental set for the given differential
equation? [Choose/Yes/No]
Solution: (a)
The wronksian of the two functions is
W(y1,y2) =
e3xe−6x
3e3x−6e−6x
=−9e−3x
(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:
•
•
(incorrect)
Correct Answers:
•-9*eˆ(-3*x)
•Yes
2
8. (1 point) It can be shown that y1=e−3xand y2=xe−3x
are solutions to the differential equation d2y
dx2+6dy
dx +9y=0 on
(−∞,∞).
(a) What does the Wronskian of y1,y2equal on (−∞,∞)?
W(y1,y2)= on (−∞,∞).
(b) Is {y1,y2}a fundamental set for the given differential
equation? [Choose/Yes/No]
Solution: (a)
The wronksian of the two functions is
W(y1,y2) =
e−3xxe−3x
−3e−3xe−3x−3xe−3x
=e−6x
(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:
•
•
(incorrect)
Correct Answers:
•eˆ(-6*x)
•Yes
9. (1 point) It can be shown that {et,tet}is a fundamental set
of solutions of y00 −2y0+y=0.
Determine which of the following is also a fundamental set.
•A. {−tet,5tet}
•B. {et−tet,−et+tet}
•C. {et+tet,et}
•D. {tet,t2et}
•E. {5et,2tet}
•F. {et−tet,et+tet}
Solution: By the principle of superposition, every linear
combination of the functions etand tetis a solution of the given
Differential Equation. To form a fundamental set, the functions
must also be linearly indendent. The sets that are formed by
functions that are both solutions and are linearly independent
are CEF
Answer(s) submitted:
•( C, E, F )
(correct)
Correct Answers:
•CEF
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