Elizabeth Arena Sharma MAT 275 ONLINE B Summer 2019
Assignment Section 2.1 Integrating Factor due 07/17/2019 at 11:59pm MST
1. (1 point) Solve the initial value problem:
dy
dx +5y=9,y(0) = 0
y(x) = .
Solution: The integrating factor is µ(x) = eR5dx =e5x.
Multiplying the differential equation by µ(x), we obtain
dy
dx e5x+5ye5x=9e5x
or d
dx ye5x=9e5x
Then, by integrating both sides of this equation, we have
ye5x=9
5e5x+C
The initial condition requires that C=−9
5. Thus
ye5x=9
5e5x−9
5
Solving for yyields the solution
y(x) = 9
5(1−e−5x)
Answer(s) submitted:
•9/5-(9eˆ(-5x)/5)
(correct)
Correct Answers:
•(9/5)*(1-eˆ(- 5*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=sinxto 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.
5tdy
dt +y=t5,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 5t:
dy
dt +1
5ty=1
5t4
The integrating factor is
µ(t) = e1
5R1
tdt =e1
5lnt=elnt1/5=t1/5.
Multiplying the differential equation by µ(t), we obtain
dy
dt t
1
5+1
5t−4
5y=1
5t
21
5
or
d
dt yt
1
5=1
5t
21
5
Then, by integrating both sides of this equation, we have
yt
1
5=1
26 t
26
5+C
Solving for yyields the general solution
y(t) = t5
26 +Ct−1
5
Answer(s) submitted:
•tˆ(1/5)
•(1/26)tˆ5+Ctˆ(-1/5)
(correct)
Correct Answers:
•tˆ(1/5)
•(tˆ5)/26+C*tˆ(-1/5)
1
4. (1 point)
Solve the following initial value problem:
tdy
dt +3y=7t
with y(1) = 2.
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 +3
ty=7
The integrating factor is
ρ(t) = e3R1
tdt =e3 lnt=elnt3=t3.
Multiplying the differential equation by ρ(t), we obtain
dy
dt t3+3t2y=7t3
or
d
dt yt3=7t3
Then, by integrating both sides of this equation, we have
yt3=7
4t2+C
Solving for yyields the general solution
y(t) = 7
4t+Ct−3
The initial condition gives C=1
4so the solution of the initial
value problem is
y(t) = 7
4t+1
4t−3
Answer(s) submitted:
•tˆ3
•(7/4)t+1/(4tˆ3)
(correct)
Correct Answers:
•tˆ(3)
•( 7 * t / (1 +3)) + 0.25 * (t**(-3 ))
5. (1 point)
Solve the initial value problem
5sin(t)dy
dt +cos(t)y=cos(t)sin7(t),
for 0 <t<πand y(π/2) = 8.
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 5sin(t):
dy
dt +cos(t)
sin(t)y=1
5cos(t)sin6(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
5cos(t)sin7(t)
or d
dt (ysin(t)) = 1
5cos(t)sin7(t)
Then, by integrating both sides of this equation, we have
ysin(t) = 1
40 sin8(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
40 sin7(t) + C
sin(t)
The initial condition gives C=319
40 so the solution of the initial
value problem is
y(t) = 1
40 sin7(t) + 319
40sin(t)
Answer(s) submitted:
•sin(t)
•(1/40)sinˆ7t+(319/(40sin(t)))
(correct)
Correct Answers:
•sin(t)
•7.975/sin(t) + 0.025 * ((sin(t))**7 )
6. (1 point)
Solve the initial value problem
6(t+1)dy
dt −5y=5t,
for t>−1 with y(0) = 13.
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=7 so the solution of the initial
value problem is
y(t) = 5t+6+7(t+1)5
6
Answer(s) submitted:
•1/(t+1)ˆ(5/6)
•5t+7(t+1)ˆ(5/6)+6
(correct)
Correct Answers:
•(t+1)ˆ(-0.833333333333333)
•5*t +6 + (7 * ((t + 1)**0.833333333333333))
7. (1 point)
Solve the following initial value problem:
dy
dt +0.5ty =3t,y(0) = 6
y(t) = .
Solution: The integrating factor is
µ(t) = e0.5Rt dt =e0.25t2
Multiplying the differential equation by µ(t), we obtain
dy
dt e0.25t2+0.5tye0.25t2=3te0.25t2
or d
dt ye0.25t2=3te0.25t2
Then, by integrating both sides of this equation, we have
ye0.25t2=3
0.5e0.25t2+C
where we have used the substitution u=0.25t2to integrate the
right hand side. Solving for yyields the general solution
y(t) = 3
0.5+Ce−0.25t2
The initial condition gives C=6−3
0.5so the solution of the
initial value problem is
y(t) = 3
0.5+6−3
0.5e−0.25t2
Answer(s) submitted:
•6
(correct)
Correct Answers:
•(3/0.5)+0*exp(-0.5*t*t/2)
8. (1 point)
Find the solution of the following IVP:
dy
dt −2ty =−3t2et2,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=−3t2
or d
dt ye−t2=−3t2
Then, by integrating both sides of this equation, we have
ye−t2=−t3+C
Solving for yyields the general solution
y(t) = −t3et2+Cet2
The initial condition gives C=−1 so the solution of the initial
value problem is
y(t) = (−t3−1)et2
Answer(s) submitted:
•eˆtˆ2(-tˆ3-1)
(correct)
Correct Answers:
•(-1*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) = e2x
Multiplying the differential equation by µ(x), we obtain
dy
dx e2x+2ye2x=−xe2x
or d
dx ye2x=−xe2x
3
Then, by integrating both sides of this equation, we have
ye2x=−x
2e2x+e2x
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+Ce−2x
Answer(s) submitted:
•eˆ(2x)
•-(x/2)+(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=4et+4e3t,y(0) = 7
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=4+4e2t
or d
dt ye−t=4+4e2t
Then, by integrating both sides of this equation, we have
ye−t=4t+2e2t+C
Solving for yyields the general solution
y(t) = 4tet+2e3t+Cet
The initial condition gives C=5. Thus the solution to the initial
value problem is
y(t) = 4tet+2e3t+5et
Answer(s) submitted:
•eˆt(4t+2eˆ(2t)+5)
(correct)
Correct Answers:
•(7 - 2 )* exp(t) + 4 * t * exp(t) + 2 * exp(3 *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+9/5)eˆ(5t)-4t-eˆt-4/5
•-9/5
(correct)
Correct Answers:
•-4*t-4*eˆt/4-4/5+eˆ(5*t)*(y0+9/5)
•-1.8
4
