Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 2.1 Integrating Factor due 03/11/2021 at 11:59pm MST
1. (1 point) Solve the initial value problem:
dy
dx +6y=3,y(0) = 0
y(x) = .
Solution: The integrating factor is µ(x) = eR6dx =e6x.
Multiplying the differential equation by µ(x), we obtain
dy
dx e6x+6ye6x=3e6x
or d
dx ye6x=3e6x
Then, by integrating both sides of this equation, we have
ye6x=1
2e6x+C
The initial condition requires that C=−1
2. Thus
ye6x=1
2e6x−1
2
Solving for yyields the solution
y(x) = 1
2(1−e−6x)
Answer(s) submitted:
•-((eˆ(-6x))/2)+(1/2)
(correct)
Correct Answers:
•(3/6)*(1-eˆ(- 6*x))
2. (1 point) Solve the initial value problem
dy
dx +ycosx=7cosx,y(0) = 9
y(x)= .
Solution: The integrating factor is µ(x) = eRcos(x)dx =esinx.
Multiplying the differential equation by µ(x), we obtain
dy
dx esinx+ycosx esinx=7cosx esinx
or d
dx yesinx=7cosx esinx
Then, by integrating both sides of this equation, we have
yesinx=7esinx+C
where we have used the substitution u=sin xto integrate the
right hand side. The initial condition requires that C=2. Thus
yesinx=7esinx+2
Solving for yyields the solution
y(x) = 7+2e−sinx
Answer(s) submitted:
•7+2eˆ(-sin(x))
(correct)
Correct Answers:
•7 + 2*eˆ(-sin(x))
3. (1 point)
Find the general solution, y(t), which solves the problem
below, by the method of integrating factors.
6tdy
dt +y=t7,t>0
Put the problem in standard form.
Then find the integrating factor, µ(t) = ,
and finally find y(t) = . (use Cas the
unkown constant.)
Solution: We put the differential equation in standard form
by dividing both sides by 6t:
dy
dt +1
6ty=1
6t6
The integrating factor is
µ(t) = e1
6R1
tdt =e1
6lnt=elnt1/6=t1/6.
Multiplying the differential equation by µ(t), we obtain
dy
dt t
1
6+1
6t−5
6y=1
6t
37
6
or
d
dt yt
1
6=1
6t
37
6
Then, by integrating both sides of this equation, we have
yt
1
6=1
43 t
43
6+C
Solving for yyields the general solution
y(t) = t7
43 +Ct−1
6
Answer(s) submitted:
•tˆ((1/6))
•(1/(43))tˆ(7)+Ctˆ(-(1/6))
(correct)
Correct Answers:
•tˆ(1/6)
•(tˆ7)/43+C*tˆ(-1/6)
1
4. (1 point)
Solve the following initial value problem:
tdy
dt +7y=7t
with y(1) = 6.
Put the problem in standard form.
Then find the integrating factor, ρ(t) = ,
and finally find y(t) = .
Solution: We put the differential equation in standard form
by dividing both sides by t:
dy
dt +7
ty=7
The integrating factor is
ρ(t) = e7R1
tdt =e7 lnt=elnt7=t7.
Multiplying the differential equation by ρ(t), we obtain
dy
dt t7+7t6y=7t7
or
d
dt yt7=7t7
Then, by integrating both sides of this equation, we have
yt7=7
8t6+C
Solving for yyields the general solution
y(t) = 7
8t+Ct−7
The initial condition gives C=41
8so the solution of the initial
value problem is
y(t) = 7
8t+41
8t−7
Answer(s) submitted:
•tˆ(7)
•(t/7)+((41)/7)Ctˆ(-7)
(score 0.5)
Correct Answers:
•tˆ(7)
•( 7 * t / (1 +7)) + 5.125 * (t**(-7 ))
5. (1 point)
Solve the initial value problem
3sin(t)dy
dt +cos(t)y=cos(t)sin2(t),
for 0 <t<πand y(π/2) = 6.
Put the problem in standard form.
Then find the integrating factor, ρ(t) = ,
and finally find y(t) = .
Solution: We put the differential equation in standard form
by dividing both sides by 3sin(t):
dy
dt +cos(t)
sin(t)y=1
3cos(t)sin1(t)
The integrating factor is
ρ(t) = eRcos(t)
sin(t)dt =eln(sin(t)) =sin(t).
Multiplying the differential equation by ρ(t), we obtain
dy
dt sin(t) + cos(t)y=1
3cos(t)sin2(t)
or d
dt (ysin(t)) = 1
3cos(t)sin2(t)
Then, by integrating both sides of this equation, we have
ysin(t) = 1
9sin3(t) +C
where we have used the substitution u=sin(t)to integrate the
right hand side.
Solving for yyields the general solution
y(t) = 1
9sin2(t) + C
sin(t)
The initial condition gives C=53
9so the solution of the initial
value problem is
y(t) = 1
9sin2(t) + 53
9sin(t)
Answer(s) submitted:
•sint
•(((6+(1/9)(sinˆ(3)(t)-1)))/(sin(t)))
(correct)
Correct Answers:
•sin(t)
•5.88888888888889/sin(t) + 0.111111111111111 * ((sin(t))**2 )
6. (1 point)
Solve the initial value problem
6(t+1)dy
dt −5y=5t,
for t>−1 with y(0) = 11.
Put the problem in standard form.
Then find the integrating factor, ρ(t) = ,
and finally find y(t) = .
Solution: We put the differential equation in standard form
by dividing both sides by 6(t+1):
dy
dt −5
6
1
t+1y=5
6
t
t+1
The integrating factor is
ρ(t) = e−5
6R1
t+1dt =e−5
6ln(t+1)= (t+1)−5
6.
Multiplying the differential equation by ρ(t), we obtain
(t+1)−5
6dy
dt −5
6(t+1)−11
6y=5
6t(t+1)−11
6
or
d
dt y(t+1)−5
6=5
6t(t+1)−11
6
2
Then, by integrating both sides of this equation, we have
y(t+1)−5
6=−t(t+1)−5
6+6(t+1)1
6+C
Solving for yyields the general solution
y(t) = −t+6(t+1) +C(t+1)5
6
The initial condition gives C=5 so the solution of the initial
value problem is
y(t) = 5t+6+5(t+1)5
6
Answer(s) submitted:
•(t+1)ˆ(-(5/6))
•((5(36t+6(t+1)ˆ((5/6))-5(t+1)ˆ((5/6))+42))/(36(t+1)ˆ((5/6))))
(score 0.5)
Correct Answers:
•(t+1)ˆ(-0.833333333333333)
•5*t +6 + (5 * ((t + 1)**0.833333333333333))
7. (1 point)
Solve the following initial value problem:
dy
dt +0.8ty =2t,y(0) = 3
y(t) = .
Solution: The integrating factor is
µ(t) = e0.8Rt dt =e0.4t2
Multiplying the differential equation by µ(t), we obtain
dy
dt e0.4t2+0.8tye0.4t2=2te0.4t2
or d
dt ye0.4t2=2te0.4t2
Then, by integrating both sides of this equation, we have
ye0.4t2=2
0.8e0.4t2+C
where we have used the substitution u=0.4t2to integrate the
right hand side. Solving for yyields the general solution
y(t) = 2
0.8+Ce−0.4t2
The initial condition gives C=3−2
0.8so the solution of the
initial value problem is
y(t) = 2
0.8+3−2
0.8e−0.4t2
Answer(s) submitted:
•(((eˆ(-((2tˆ(2))/5))))/2)+(((5))/2)
(correct)
Correct Answers:
•(2 / 0.8 ) + 0.5 * exp( -0.8 * t * t /2)
8. (1 point)
Find the solution of the following IVP:
dy
dt −2ty =9t2et2,y(0) = 1.
y(t) = .
Solution: The integrating factor is
µ(t) = eR(−2t)dt =e−t2
Multiplying the differential equation by µ(t), we obtain
dy
dt e−t2−2tye−t2=9t2
or d
dt ye−t2=9t2
Then, by integrating both sides of this equation, we have
ye−t2=3t3+C
Solving for yyields the general solution
y(t) = 3t3et2+Cet2
The initial condition gives C=1 so the solution of the initial
value problem is
y(t) = (3t3+1)et2
Answer(s) submitted:
•3tˆ(3)eˆ(tˆ(2))+eˆ(tˆ(2))
(correct)
Correct Answers:
•(3*tˆ3 + 1)*eˆ(tˆ2)
9. (1 point) Find the general solution to the differential equa-
tion
x2−2xy +xdy
dx =0
Put the problem in standard form.
Find the integrating factor, ρ(x) = .
Find y(x) = .
Use Cas the unknown constant.
Solution: We convert the problem in standard form by divid-
ing by xand rearranging terms:
dy
dx −2y=−x
The integrating factor is
µ(x) = e−2x
Multiplying the differential equation by µ(x), we obtain
dy
dx e−2x−2ye−2x=−xe−2x
or d
dx ye−2x=−xe−2x
3
Then, by integrating both sides of this equation, we have
ye−2x=x
2e−2x+e−2x
4+C
where we have used integration by parts to integrate the right
hand side. Solving for yyields the general solution
y(t) = x
2+1
4+Ce2x
Answer(s) submitted:
•eˆ(-2x)
•(((2x+1))/4)+Ceˆ(2x)
(correct)
Correct Answers:
•eˆ(-2 x)
•C eˆ(2 x) + x/2 + 1/2ˆ2
10. (1 point)
Solve the initial value problem
dy
dt −y=3et+18e4t,y(0) = 9
y(t) = .
Solution: The integrating factor is
µ(t) = e−t
Multiplying the differential equation by µ(t), we obtain
dy
dt e−t−ye−t=3+18e3t
or d
dt ye−t=3+18e3t
Then, by integrating both sides of this equation, we have
ye−t=3t+6e3t+C
Solving for yyields the general solution
y(t) = 3tet+6e4t+Cet
The initial condition gives C=3. Thus the solution to the initial
value problem is
y(t) = 3tet+6e4t+3et
Answer(s) submitted:
•3eˆ(t)t+6eˆ(4t)+3eˆ(t)
(correct)
Correct Answers:
•(9 - 6 )* exp(t) + 3 * t * exp(t) + 6 * exp(4 *t)
11. (1 point) Consider the initial value problem
y0−5y=20t+4et,y(0) = y0
(a) Solve the initial value problem. (enter y0 for y0).
y(t)=
(b) Determine the value of y0that separates solutions that
grow positively as t→∞from those that grow negatively.
y0=
Solution: (a)
The integrating factor is ρ(t) = eR(−5)dt =e−5t.
Multiplying both sides of the differential equation by ρ(t)yields
d
dt e−5ty=20te−5t+4e−4t
Integrating both sides yields:
e−5ty=Z20te−5t+4e−4tdt
e−5ty=−4t−4
5e−5t−1e−4t+C
Solving or y:
y=−4t−4
5−1et+Ce5t
Subsstituting the initial condition y(0) = y0, gives C=y0+9
5
and the solution is
y=−4t−4
5−1et+y0+9
5e5t
(b)
As t→∞, the last term is the dominant term. Thus the value of
y0that separates solutions that grow positively from those that
grow negatively is
y0=−9
5
Answer(s) submitted:
•(y0+(5/(20)))eˆ(-5t)-4t-(4/5)-2eˆ(t)
•-((28)/(20))
(incorrect)
Correct Answers:
•-4*t-4*eˆt/4-4/5+eˆ(5*t)*(y0+9/5)
•-1.8
4
