Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 3.4 Repeated Roots due 03/24/2021 at 11:59pm MST
1. (1 point)
Find the general solution to the homogeneous differential
equation.
d2y
dt2+8dy
dt +16y=0
Use c1and c2in your answer to denote arbitrary constants, and
enter them as c1 and c2.
y(t) = help (formulas)
Answer(s) submitted:
•c1eˆ(-4t)+c2teˆ(-4t)
(correct)
Correct Answers:
•c1*eˆ(-4*t)+c2*t*eˆ(-4*t)
2. (1 point)
Find yas a function of tif
16y00 −136y0+289y=0,
y(0) = 2,y0(0) = 7.
y(t) =
Solution: The characteristic equation is
16r2−136r+289 = (4r−17)2=0.
Thus the polynomial has the repeated root r=17
4and the general
solution can be written as
y=c1e
17
4t+c2te
17
4t
,
for which y0=17
4c1e
17
4t+c2e
17
4t+17
4c2te
17
4t.
Substituting the initial conditions gives the system
2=c1
7=17
4c1+c2
with solution c1=2,c2=−3
2.
Hence the solution to the initial value problem is
y=2e
17
4t−3
2te
17
4t
Answer(s) submitted:
•2eˆ(((17t)/4))-((3eˆ(((17t)/4))t)/2)
(correct)
Correct Answers:
•(2) *exp((17/4)*t) + (-3/2)*t *exp((17/4)*t)
3. (1 point)
Find yas a function of tif
9y00 +24y0+16y=0,
y(0) = 5,y0(0) = 6.
y(t) =
Solution: The characteristic equation is
9r2+24r+16 = (3r+4)2
.
Thus the polynomial has the repeated root r=−4
3and the gen-
eral solution can be written as
y=c1e−4
3t+c2te−4
3t
,
for which y0=−4
3c1e−4
3t+c2e−4
3t−4
3c2te−4
3t.
Substituting the initial conditions gives the system
5=c1
6=−4
3c1+c2
with solution c1=5,c2=38
3.
Hence the solution to the initial value problem is
y=5e−4
3t+38
3te−4
3t
Answer(s) submitted:
•5eˆ(-((4t)/3))+((38eˆ(((-4t)/3))t)/3)
(correct)
Correct Answers:
•(5) *exp((-4/3)*t) + (38/3)*t *exp((-4/3)*t)
4. (1 point) Match the third order linear equations with their
fundamental solution sets.
1. y000 −y00 −y0+y=0
2. y000 +3y00 +3y0+y=0
3. ty000 −y00 =0
4. y000 −6y00 +y0−6y=0
5. y000 −8y00 +15y0=0
6. y000 +y0=0
A. et
,tet
,e−t
B. e6t
,cos(t),sin(t)
C. 1,t,t3
D. e−t
,te−t
,t2e−t
E. 1,cos(t),sin(t)
F. 1,e5t
,e3t
Answer(s) submitted:
•a
•d
1
•c
•b
•f
•e
(correct)
Correct Answers:
•A
•D
•C
•B
•F
•E
5. (1 point)
Find the solution to initial value problem
d2y
dt2+14 dy
dt +49y=0,y(0) = 3,y0(0) = 5
y(t) = .
Solution: The characteristic equation is
r2+14r+49 = (r+7)2=0.
Thus the polynomial has the repeated root r=−7 and the gen-
eral solution can be written as
y=c1e−7t+c2te−7t
,
for which y0=−7c1e−7t+c2e−7t−7c2te−7t.
Substituting the initial conditions gives the system
3=c1
5=−7c1+c2
with solution c1=3,c2=26.
Hence the solution to the initial value problem is
y=3e−7t+26te−7t
Answer(s) submitted:
•3eˆ(-7t)+26teˆ(-7t)
(correct)
Correct Answers:
•3*exp(-7*t) + 26*t*exp(-7*t)
6. (1 point)
Find yas a function of xif
y000 −11y00 +18y0=0,
y(0) = 8,y0(0) = 9,y00 (0) = 3.
y(x) =
Solution: The characteristic equation is
r3−11r2+18r=r(r2−11r+18) = r(r−2)(r−9) = 0,
with roots r1=0,r2=2 and r3=9. The general solution can
be written as
y=c1+c2e2t+c3e9t
,
for which y0=2c2e2t+9c3e9tand y00 =4c2e2t+81c3e9t.
Substituting the initial conditions gives the system
8=c1+c2+c3
9=2c1+9c2
3=4c1+81c2
with solution c1=8
3,c2=39
7and c3=−5
21 .
Hence the solution to the initial value problem is
y=2.66667 +5.57143e2x−0.238095e9x
Answer(s) submitted:
•8-((39)/7)+(5/(21))+((39eˆ(2x))/7)-(5/(21))eˆ(9x)
(correct)
Correct Answers:
•2.66666666666667 + 5.57142857142857*eˆ(2*x) + -0.238095238095238*eˆ(9*x)
7. (1 point)
Find yas a function of xif
y(4)−12y000 +36y00 =0,
y(0) = 6,y0(0) = 3,y00 (0) = 36,y000 (0) = 0.
y(x) =
Solution: The characteristic equation is
r4−12r3+36r2=r2(r2−12r+36) = r(r−6)2
,
with roots r1=0 (repeated) and r2=6( repeated). The general
solution can be written as
y=c1+c2x+c3e6x+c4xe6x
,
for which
y0=c2+6c3e6x+c4e6x+6c4xe6x
y00 =36c3e6x+12c4e6x+36c4xe6x
y000 =216c3e6x+108c4e6x+216c4xe6x
Substituting the initial conditions gives the system
6=c1+c3
3=c2+6c3+c4
36 =36c3+12c4
0=216c3+108c4
with solution c1=3,c2=−9,c3=3 and c4=−6.
Hence the solution to the initial value problem is
y=3−9x+3e6x−6xe6x
Answer(s) submitted:
•-9x-6eˆ(6x)x+3eˆ(6x)+3
(correct)
Correct Answers:
•3 + -9*x + 3*eˆ(6*x) + -6*x*eˆ(6*x)
2
8. (1 point)
Find yas a function of xif
y000 −5y00 −y0+5y=0,
y(0) = −8,y0(0) = −6,y00 (0) = −8.
y(x) =
Solution: The characteristic equation is
r3−5r2−r+5=r2(r−5)−(r−5) = (r2−1)(r−5) = 0,
with roots r1=1,r2=−1 and r3=5. The general solution can
be written as
y=c1ex+c2e−x+c3e5x
,
for which y0=c1ex−c2e−x+5c3e5xand y00 =c1ex+c2e−x+
25c3e5x.
Substituting the initial conditions gives the system
−8=c1+c2+c3
−6=c1−c2+5c3
−8=c1+c2+25c2
with solution c1=−7,c2=−1 and c3=0.
Hence the solution to the initial value problem is
y=−7ex−e−x
Answer(s) submitted:
•-7eˆ(x)-eˆ(-x)
(correct)
Correct Answers:
•-7*eˆx + -1*eˆ(-x) + 0*eˆ(5*x)
9. (1 point) The differential equation
x2d2y
dx2−7xdy
dx +16y=0
has x4as a solution.
Applying reduction of order we set y2=ux4.
Then (using the prime notation for the derivatives)
y0
2=
y00
2=
So, plugging y2into the left side of the differential equation,
and reducing, we get
x2y00
2−7xy0
2+16y2=
The reduced form has a common factor of x5which we can
divide out of the equation so that we have xu00 +u0=0.
Since this equation does not have any u terms in it we can make
the substitution w=u0giving us the first order linear equation
xw0+w=0.
This equation has integrating factor for x ¿ 0.
If we use a as the constant of integration, the solution to this
equation is w=
Integrating to get u, and using b as our second constant of inte-
gration we have u=
Finally y2=and the general solution is
Answer(s) submitted:
•
•
•
•
•
•
•
•
(incorrect)
Correct Answers:
•4*xˆ3*u+xˆ4*u’
•12*xˆ2*u+8*xˆ3*u’+xˆ4*u’’
•xˆ6*u’’+xˆ5*u’
•x
•a/x
•a*ln(x)+b
•a*xˆ4*ln(x)+b*xˆ4
•a*xˆ4*ln(x)+b*xˆ4
10. (1 point) The differential equation
t2y00 −t(t+2)y0+ (t+2)y=0
has y1=tas a solution.
Applying reduction of order we set y2=v·y1=v·t.
Then (using the prime notation for the derivatives)
y0
2=
y00
2=
So, substituting y2and its derivatives into the left side of the
differential equation, and reducing, we get
t2y00
2−t(t+2)y0
2+ (t+2)y2=
The reduced form has a common factor of t3which we can
divide out of the equation.
Since this equation does not have any vterms in it we can make
the substitution u=v0giving us the first order linear equation in
u:
=0.
If we use c as the constant of integration, the solution to this
equation is u=
Integrating to get v, and then finding y2gives the general solu-
tion:
c1y1+c2y2=
Answer(s) submitted:
•
•
•
•
•
•
3
(incorrect)
Correct Answers:
•v+v’*t
•v’’*t+2*v’
•v’’*tˆ3-v’*tˆ3
•u’-u
•c*eˆt
•c1*t+c2*t*eˆt
4
