Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 3.5 Undetermined Coefficients due 03/24/2021 at 11:59pm MST
1. (1 point) Use the method of undetermined coefficients to
find one solution of
y00 −6y0+11y=7e5t
.
y(t) =
Solution: The complementary solution is
yc=c1e6+√−8
2t+c2e6−√−8
2t
Since none of the terms in ycduplicates e5t, we assume the par-
ticular solution has the form
y(t) = Ae5t
Substituting y(t) = Ae5t,y0(t) = 5Ae5tand y00(t) = 25Ae5tin the
differential equation yields
25Ae5t−6·5Ae5t+11 ·5Ae5t=7e5t
,
which simplifies to
6Ae5t=7e5t
.
Hence A=7
6and a particular solution is
y(t) = 7
6e5t
Answer(s) submitted:
•(7/6)eˆ(5t)
(correct)
Correct Answers:
•( 7/6 ) * exp((5)*t) + c*eˆ(3*t)cos(1.4142135623731*t) +d*eˆ(3*t)*sin(1.4142135623731*t)
2. (1 point)
Find a particular solution to the differential equation
−6y00 −y0+y=t2+2t+2e4t
.
yp=
Solution: The complementary solution is yc=c1e0.333333t+
c2e−t
2.
Since none of the terms in t2+2t+2e4tand its derivatives dupli-
cates a term in yc, we assume a particular solution of the form:
yp=A+Bt +Ct2+De4t
,
for which
y0p=B+2Ct +4De4tand y00
p=2C+16De4t
.
Substitution in the differential equation yields
−62C+16De4t+B+2Ct +4De4t+A+Bt +Ct2+De4t=t2+2t+2e4t
,
which simlifies to
(A−(12C+B)) + (B−2C)t+Ct2−99De4t=t2+2t+2e4t
.
We equate coefficients of like terms to get the system
A−(12C+B) = 0
B−2C=2
C=1
−99D=2
with solution A=16,B=4,C=1 and D=−2
99 .
Hence a particular solution is
yp=16 +4t+1t2−2
99 e4t
Answer(s) submitted:
•tˆ(2)+4t+16-(2/(99))eˆ(4t)
(correct)
Correct Answers:
•1*(t**2) + 4 *(t) + 16 + -0.0202020202020202 exp(4*t) + a*exp(0.333333333333333*t) + b*(exp(-0.5*t) + 0*t*exp(-0.5*t) )
3. (1 point)
Find a particular solution to
y00 −2y0+y=−14.5et
.
yp=
Solution: The complementary solution is yc=c1et+c2tet.
To eliminate duplication, we assume the particular solution has
the form
yp=At2et
,
for which
y0p=2Atet+At2etand y00
p=2Aet+4Atet+At2et
.
Substituting in the original differential equation and simplifying
yields
2Aet=−14.5et
Thus A=−7.25 and a particular solution is
yp=−7.25t2et
Answer(s) submitted:
•-((29eˆ(t)tˆ(2))/4)
(correct)
Correct Answers:
•(-14.5/2)*(t**2)*exp(1*t) + a*eˆ(1*t) + b*t*e**(1*t)
1
4. (1 point)
Find a particular solution to the differential equation
y00 −5y0+4y=−32t3
.
yp=
Solution: The complementary solution is yc=c1e4t+c2et.
Since t3and its derivatives do not duplicates any of the terms in
yc, we assume a particular solution of the form
yp=A+Bt +Ct2+Dt3
,
for which
y0p=B+2Ct +3Dt2and y00
p=2C+6Dt.
Substituton in the original differential equation yields
2C+6Dt −5B+2Ct +3Dt2+4A+Bt +Ct2+Dt3=−32t3
Rearranging terms gives
(2C−5B+4A)+(6D−10C+4B)t+(4C−15D)t2+4D=−32t3
We equate coefficients of like powers of tto get
2C−5B+4A=0
6D−10C+4B=0
4C−15D=0
4D=−32
with solution A=−255
4,B=−63,C=−30 and D=−8.
Hence a particular solution is
yp=−255
4−63t−30t2−8t3
Answer(s) submitted:
•-8tˆ(3)-30tˆ(2)-63t-((255)/4)
(correct)
Correct Answers:
•-8*(t**3) + -30 *(t**2) + -63*t + -63.75 + a*eˆ(4*t) + b*(eˆ(1*t) + 0*t*e**(4*t))
5. (1 point)
Find a particular solution to
y00 +4y0+3y=−12te5t
.
yp=
Solution: The complementary solution is yc=c1e−t+
c2e−3t.
Since te5tand its derivatives do not duplicate any of the terms
in yc, we assume a particular solution of the form:
yp= (A+Bt)e5t
,
for which
y0p= (5Bt +5A+B)e5tand y00
p= (25Bt +25A+10B)e5t
.
Substitution in the original differential equation yields
(25Bt +25A+10B)e5t+4(5Bt +5A+B)e5t+3(A+Bt)e5t=−12te5t
,
which simplifies to
(48A+14B) + 48Bt =−12t
We equate coefficients to get the system
48A+14B=0
48B=−12
with solution A=7
96 and B=−1
4.
Thus a particular solution is
yp=7
96 −1
4te5t
Answer(s) submitted:
•-((eˆ(5t)t)/4)+(7/(96))eˆ(5t)
(correct)
Correct Answers:
•(-0.25 * t + 0.0729166666666667) * ((2.71828182845905)**(5*t)) + a*eˆ(-1*t) + b*e**(-3*t)
6. (1 point)
(1) Find a particular solution to the nonhomogeneous dif-
ferential equation y00 +4y0+5y=−10x+5e−x.
yp=help (formulas)
(2) Find the complementary solution. Use c1and c2in your
answer to denote arbitrary constants, and enter them as
c1 and c2.
yc=help (formulas)
(3) Find the most general solution to the original nonho-
mogeneous differential equation. Use c1and c2in your
answer to denote arbitrary constants.
y=help (formulas)
Solution: a.
Based on the form of the right hand side we assume a particular
solution of the form yp=Ax +B+Ce−x.
We check for duplication with the homogeneous. The character-
istic equation r2+4r+5=0 has roots r1,2=−2±i. Thus the
complementary solution is yc=c1e−2xcos(x) + c2e−2xsin(x)
and yp=Ax +B+Ce−xis the correct form of the particular
solution.
Substituting yp,y0p=A−Ce−xand y00
p=Ce−xin the differential
equation yields
Ce−x+4(A−Ce−x) + 5(Ax +B+Ce−x) = −10x+5e−x
Rearring terms gives
2Ce−x+5Ax + (4A+5B) = −10x+5e−x
Matching coefficients of like terms yields the system
2C=5
5A=−10
4A+5B=0
2
with solution A=−2,B=8
5and C=5
2.
Thus the particular solution is
yp=−2x+8
5+5
2e−x
b. The complementary solution was already found in part a.
yc=c1e−2xcos(x) + c2e−2xsin(x)
c. The general solution is the sum of the complementary and the
particular:
y=yc+yp=c1e−2xcos(x) + c2e−2xsin(x)−2x+8
5+5
2e−x
Answer(s) submitted:
•-2x+(8/5)+(5/2)eˆ(-x)
•eˆ(-2x)(c1cos(x)+c2sin(x))
•eˆ(-2x)(c1cos(x)+c2sin(x))-2x+(8/5)+(5/2)eˆ(-x)
(correct)
Correct Answers:
•-4*-2/5+-2*x+2.5*eˆ(-x)+a*eˆ(-2*x)*cos(x)+b*eˆ(-2*x)*sin(x)
•c1*eˆ(-2*x)*cos(x)+c2*eˆ(-2*x)*sin(x)
•-4*-2/5+-2*x+2.5*eˆ(-x)+c1*eˆ(-2*x)*cos(x)+c2*eˆ(-2*x)*sin(x)
7. (1 point)
Find the solution of
y00 +13y0+42y=36e−3t
with y(0) = 7 and y0(0) = 9.
y=
Solution: The complementary solution is yc=c1e−7t+
c2e−6t.
Since e−3tand its derivatives do not duplicate any of the terms
in yc, we assume a particular solution of the form
yp=Ae−3t
,
for which
y0p=−3Ae−3tand y00
p=9Ae−3t
.
Substitution in the original differential equation yields
9Ae−3t−39Ae−3t+42Ae−3t=36e−3t
,
which simplifies to 12A=36.Thus A=3 and a particular solu-
tion is
yp=3e−3t
.
The general solution is
y=c1e−7t+c2e−6t+3e−3t
Substituting the initial conditions yields the system
7=c1+c2+3
9=−7c1−6c2−9
with solution c1=−42 and c2=46.
Thus the solution to the initial value problem is
y=−42e−7t+46e−6t+3e−3t
Answer(s) submitted:
•46eˆ(-6t)-42eˆ(-7t)+3eˆ(-3t)
(correct)
Correct Answers:
•( 3 ) * exp((-3)*t) + (2 - 44) *exp((-13/2 - 1/2)*t) + (2 + 44) *exp((-13/2 + 1/2)*t)
8. (1 point)
Find the solution of
y00 +2y0=16 sin(2t) + 40 cos(2t)
with y(0) = 4 and y0(0) = 7.
y=
Solution: The complementary solution is yc=c1+c2e−2t.
Since sin(2t)and cos(2t)do not duplicate any of the terms in
yc, we assume a particular solution of the form
yp=Acos(2t) + Bsin(2t),
for which
y0p=−2Asin(2t)+2Bcos(2t)and y00
p=−4Acos(2t)−4Bsin(2t).
Substitution in the original differential equation yields
−4Acos(2t)−4Bsin(2t)−4Asin(2t)+4Bcos(2t) = 16 sin(2t)+40 cos(2t),
which simplifies to
4(B−A)cos(2t)−4(A+B)sin(2t) = 16 sin(2t) + 40 cos(2t).
We equate coefficients of like terms to get
4(B−A) = 40
−4(A+B) = 16
with solution A=−7 and B=3. Hence the particular solu-
tion is
yp=−7cos(2t) + 3sin(2t).
The general solution is
y=c1+c2e−2t−7cos(2t) + 3sin(2t)
Substituting the initial conditions yields the system
4=c1+c2−7
7=−2c2+6
with solution c1=23
2and c2=−1
2.
Thus the solution to the initial value problem is
y=23
2−1
2e−2t−7cos(2t) + 3sin(2t)
Answer(s) submitted:
•(1/2)+11-(1/2)eˆ(-2t)+3sin(2t)-7cos(2t)
(correct)
Correct Answers:
•(-7 )*1*cos(2 *t) + (3 )*1*sin(2 *t) + (11/2 - 6) *exp((-1 - 1)*t) + (11/2 + 6) *exp((-1 + 1)*t)
3
9. (1 point)
Find a particular solution to
y00 +9y=−18sin(3t).
yp=
Solution: The complementary solution is yc=c1cos(3t) +
c2sin(3t).
To avoid duplication, we assume a particular solution of the
form
yp=t(Acos(3t) + Bsin(3t)),
for which
y0p=Acos(3t) + Bsin(3t) + t(−3Asin(3t) + 3Bcos(3t)) and
y00
p=−6Asin(3t) + 6Bcos(3t) +t(−9Acos(3t)−9Bsin(3t)).
Substitution in the original differential equation yields
−6Asin(3t)+6Bcos(t)+t(−9Acos(3t)−9Bsin(3t))+9t(Acos(3t)+Bsin(3t)) = −18sin(3t),
which simplifies to
−6Asin(3t) + 6Bcos(t) = −18sin(3t).
Thus −6A=−18 and 6B=0, which gives A=3 and B=0.
Hence a particular solution is
yp=3tcos(3t).
Answer(s) submitted:
•3tcos(3t)
(correct)
Correct Answers:
•3 * t * cos(3*t) + a*sin(3*t) + b*cos(3*t)
10. (1 point)
Find yas a function of xif
y000 −13y00 +42y0=60ex
,
y(0) = 30,y0(0) = 23,y00(0) = 27.
y(x) =
Solution: The complementary solution is yc=c1+c2e6x+
c3e7x.
Since there is no duplication with the right hand side, we assume
a particular solution of the form
yp=Aex
,
for which y0p=Aex
,y00
p=Aex
,and y000
p=Aex.
Substitution in the original differential equation yields
Aex−13Aex+42Aex=60ex
,
which simplifies to
30Aex=60ex
.
Thus A=2 and the particular solution is
yp=2ex
.
The general solution is
y=c1+c2e6x+c3e7x+2ex
Substituting the initial conditions yields the system
30 =c1+c2+c3+2
23 =6c2+7c3+2
27 =36c2+49c3+2
with solution c1=464
21 ,c2=61
3and c3=−101
7.
Hence the solution to the initial value problem is
y(x) = 464
21 +61
3e6x−101
7e7x+2ex
Answer(s) submitted:
•28-((61)/3)+((101)/7)+((61eˆ(6x))/3)-((101)/7)eˆ(7x)+2eˆ(x)
(correct)
Correct Answers:
•22.0952380952381 + 20.3333333333333*eˆ(6*x) + -14.4285714285714*eˆ(7*x) + 2 *eˆx
11. (3 points) Match the following guess solutions ypfor
the method of undetermined coefficients with the second-order
nonhomogeneous linear equations below.
A. yp(x) = Ax2+Bx +C
B. yp(x) = Ae2x
C. yp(x) = Acos2x+Bsin2x
D. yp(x) = (Ax +B)cos2x+ (Cx +D)sin2x
E. yp(x) = Axe2x
,
F. yp(x) = e3x(Acos2x+Bsin 2x)
1. d2y
dx2−5dy
dx +6y=e2x
2. d2y
dx2+4y=−3x2+2x+3
3. y00 +4y0+20y=−3sin2x
4. y00 −2y0−15y=e3xcos2x
Solution: 1. The complementary solution is yc=c1e3x+
c2e2x. Based on the form of the right hand side we guess a par-
ticular solution of the form Ae2x. Since there is duplication with
yc, we have yp=Axe2x. Thus the answer is E.
2. The complementary solution is yc=c1cos(2x) +
c2sin(2x). Based on the form of the right hand side we guess
a particular solution of the form Ax2+Bx +Cand since there
is no duplication with yc, we have yp=Ax2+Bx +C. Thus the
answer is A.
3. The complementary solution is yc=c1e−2xcos4x+
c2e−2xsin4x. Based on the form of the right hand side we guess
a particular solution of the form Acos2x+Bsin2xand since
there is no duplication with yc, we have yp=Acos 2x+Bsin 2x.
Thus the answer is C.
4. The complementary solution is yc=c1e5x+c2e−3x.
Based on the form of the right hand side we guess a particu-
lar solution of the form Ae3xsin2x+Be3xcos2xand since there
is no duplication with yc, we have yp=Ae3xsin2x+Be3xcos2x.
Thus the answer is F.
Answer(s) submitted:
•b
•a
•c
4
•f
(incorrect)
Correct Answers:
•E
•A
•C
•F
12. (1 point) Match the following nonhomogeneous linear
equations with the form of the particular solution ypfor the
method of undetermined coefficients.
? 1. y00 +2y0+2y=3e−t+2e−tcost+4e−tt2sint
? 2. y00 −4y0+4y=2t2+4te2t+tsin(2t)
? 3. y00 +y=t(1+sint)
? 4. y00 +4y=t2sin(2t)+(5t−7)cos(2t)
A. yp=A0t2+A1t+A2+t2(B0t+B1)e2t+ (C0t+
C1)sin(2t)+(D0t+D1)cos(2t)
B. yp=Ae−t+t(B0t2+B1t+B2)e−tcost+t(C0t2+C1t+
C2)e−tsint
C. yp=t(A0t2+A1t+A2)sin(2t) + t(B0t2+B1t+
B2)cos(2t)
D. yp=A0t+A1+t(B0t+B1)sint+t(C0t+C1)cost
Solution:
SOLUTION
1. The complementary solution is yc=c1e−tcost+
Ae−t+t(B0t2+B1t+B2)e−tcost+t(C0t2+C1t+C2)e−tsint.
Thus the answer is B.
2. The complementary solution is yc=c1e2t+c2te2t.
Based on the form of the right hand side we guess a particu-
lar solution of the form A0t2+A1t+A2+(B0t+B1)e2t+(C0t+
C1)sin(2t)+(D0t+D1)cos(2t). However, the terms B1e2tand
B0te2tduplicate the homogeneous. Thus we need to multiply by
t2and the correct particular solution is yp=A0t2+A1t+A2+
t2(B0t+B1)e2t+ (C0t+C1)sin(2t) + (D0t+D1)cos(2t). Thus
the answer is A.
3. The complementary solution is yc=c1cost+c2sint.
Based on the form of the right hand side we guess a particular
solution of the form A0t+A1+(B0t+B1)sint+(C0t+C1)cost.
However, the terms B1sintand C1costduplicate the homoge-
neous. Thus we need to multiply by tand the correct particular
solution is yp=A0t+A1+t(B0t+B1)sint+t(C0t+C1)cost.
Thus the answer is D.
4. The complementary solution is yc=c1cos2t+c2sin2t.
Based on the form of the right hand side we guess a particular
solution of the form (A0t2+A1t+A2)sin(2t) + (B0t2+B1t+
B2)cos(2t). However, the terms A2sin 2tand B2cos2tduplicate
the homogeneous. Thus we need to multiply by tand the cor-
rect particular solution is yp=yp=t(A0t2+A1t+A2)sin(2t)+
t(B0t2+B1t+B2)cos(2t). Thus the answer is C.
Answer(s) submitted:
•C
•B
•D
•A
(incorrect)
Correct Answers:
•B
•A
•D
•C
c2e−t sint. Based on the form of the right hand side we
guess a particular solution of the form A0e−t + (B0t2 + B1t +
B2)e−t cost + (C0t2 + C1t + C2)e−t sint. However, the terms
B2e−t cost and C2e−t cost duplicate the homogeneous. Thus we
need to multiply by t and the correct particular solution is yp =
5
