Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 6.2 Solution of I.V.P due 03/31/2021 at 11:59pm MST
1. (1 point)
Find the inverse Laplace transform f(t) = L−1{F(s)}of
F(s) = 9s+4
s2+19 s>0
f(t) = .
Solution:
L−1n9s+4
s2+19 o=9L−1ns
s2+19 o+4
√19 L−1n√19
s2+19 o
=9cos(√19t) + 4
√19 sin(√19t)
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•9*cos(4.35889894354067*t)+0.917662935482247*sin(4.35889894354067*t)
2. (1 point) Find the inverse Laplace transform f(t) =
L−1{F(s)}of
F(s) = 2s+9
s2−36 s>6
f(t) = .
Solution:
L−1n2s+9
s2−36 o=2L−1ns
s2−36 o+3
2L−1n6
s2−36 o
=2cosh(6t) + 3
2sinh(6t)
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•2*cosh(6*t)+1.5*sinh(6*t)
3. (1 point)
Use the Laplace transform to solve the following initial value
problem:
y00 +5y0−36y=0y(0) = 4,y0(0) = −2
(1) First, using Yfor the Laplace transform of y(t), i.e.,
Y=L{y(t)}, find the equation you get by taking the
Laplace transform of the differential equation
=0
(2) Now solve for
Y(s) =
(3) Write the above answer in its partial fraction decompo-
sition, Y(s) = A
s+a+B
s+bwhere a<b
Y(s) = +
(4) Now by inverting the transform, find
y(t) = .
Solution:
(1) Taking the Laplace transform of both sides of the dif-
ferential equation yields:
s2Y−sy(0)−y0(0) + 5(sY −y(0)) −36Y=0
Substituting the initial conditions gives:
s2Y−4s+2+5sY −20 −36Y=0
(2) Solving for Ygives:
Y=4s+18
s2+5s−36 =4s+18
(s−4)(s+9)
(3) Performing Partial Fraction Decomposition yields:
4s+18
(s−4)(s+9)=
34
13
s−4+
18
13
s+9
(4) Inverting the transform, we find
y(t) = 34
13 e4t+18
13 e−9t
Answer(s) submitted:
•
•
•
•
•
(incorrect)
Correct Answers:
•sˆ2*Y-4*s--2+5*(s*Y-4)+-36*Y
•(4*s+18)/(sˆ2+5*s+-36)
•2.61538461538462/(s-4)
•1.38461538461538/(s+9)
•2.61538461538462*exp(4*t)+1.38461538461538*exp(-9*t)
4. (1 point)
Use the Laplace transform to solve the following initial value
problem:
y00 +2y0=0y(0) = 1,y0(0) = 7
(1) Using Yfor the Laplace transform of y(t), i.e., Y=
L{y(t)}, find the equation you get by taking the
Laplace transform of the differential equation
=0
(2) Now solve for
Y(s) =
(3) Write the above answer in its partial fraction decompo-
sition, Y(s) = A
s+a+B
s+bwhere a<b
Y(s) = +
1
(4) Now by inverting the transform, find
y(t) = .
Solution:
(1) Taking the Laplace transform of both sides of the dif-
ferential equation yields:
s2Y−sy(0)−y0(0) + 2(sY −y(0)) = 0
Substituting the initial conditions gives:
s2Y−s−7+2sY −2=0
(2) Solving for Ygives:
Y=s+9
s2+2s=s+9
s(s+2)
(3) Performing Partial Fraction Decomposition yields:
s+9
(s)(s+2)=
9
2
s+−7
2
s+2
(4) Inverting the transform, we find
y(t) = 9
2−7
2e−2t
Answer(s) submitted:
•
•
•
•
•
(incorrect)
Correct Answers:
•sˆ2*Y-1*s-7+2*(s*Y-1)
•(1*s+9)/(sˆ2+2*s)
•4.5/s
•-3.5/(s+2)
•4.5+-3.5*exp(-2*t)
5. (1 point)
Use the Laplace transform to solve the following initial value
problem:
y00 +9y=cos(5t)y(0) = 0,y0(0) = 0
First, using Yfor the Laplace transform of y(t), i.e., Y=
L{y(t)},
find the equation you get by taking the Laplace transform of the
differential equation and solving for Y:
Y(s) =
Find the partial fraction decomposition of Y(s)and its inverse
Laplace transform to find the solution of the IVP:
y(t) = .
Solution: Taking the Laplace transform of both sides of the
differential equation yields:
s2Y−sy(0)−y0(0) + 9Y=s
s2+25
Substituting the initial conditions gives:
s2Y+9Y=s
s2+25
Solving for Ygives:
Y=s
(s2+25)(s2+9)
We set up the Partial Fraction Decomposition:
s
(s2+25)(s2+9)=As +B
s2+25 +Cs +D
s2+9
s= (As +B)(s2+9)+(Cs +D)(s2+25)
s=s3(A+C) + s2(B+D) + s(9A+25C)+(9B+25D)
Matching powers of syields the system:
A+C=0
B+D=0
9A+25C=1
9B+25D=0
Solving the system yields B=D=0, A=−1
16 and C=1
16 .
Thus
Y=s
(s2+25)(s2+9)=−1
16
s
s2+25 +1
16
s
s2+9
Inverting the transform, we find
y(t) = −1
16 cos(5t) + 1
16 cos(3t)
Answer(s) submitted:
•
•
(incorrect)
Correct Answers:
•s/((sˆ2+9)*(sˆ2+5**2))
•1/-16*(cos(5*t)-cos(3*t))
6. (1 point)
Consider the initial value problem
y00 +16y=48t,y(0) = 8,y0(0) = 9.
(1) Take the Laplace transform of both sides of the given
differential equation to create the corresponding alge-
braic equation. Denote the Laplace transform of y(t)
by Y(s). Do not move any terms from one side of the
equation to the other (until you get to part (b) below).
=help
(formulas)
(2) Solve your equation for Y(s).
Y(s) = L{y(t)}=
(3) Take the inverse Laplace transform of both sides of the
previous equation to solve for y(t).
y(t) =
Answer(s) submitted:
2
•
•
•
•
(incorrect)
Correct Answers:
•sˆ2*Y(s)-8*s-9+16*Y(s)
•48/(sˆ2)
•48/[sˆ2*(sˆ2+16)]+(8*s+9)/(sˆ2+16)
•3*t-0.75*sin(4*t)+8*cos(4*t)+2.25*sin(4*t)
7. (1 point) Use the Laplace transform to solve the following
initial value problem:
y00 −2y0+10y=0y(0) = 0,y0(0) = 3
(1) Using Yfor the Laplace transform of y(t), i.e., Y=
L{y(t)}, find the equation you get by taking the
Laplace transform of the differential equation
=0
(2) Now solve for
Y(s) =
(3) By completing the square in the denominator and in-
verting the transform, find
y(t) = .
Solution:
(1) Taking the Laplace transform of both sides of the dif-
ferential equation yields:
s2Y−sy(0)−y0(0)−2(sY −y(0)) + 10Y=0
Substituting the initial conditions gives:
s2Y−3−2sY +10Y=0
(2) Y=3
s2−2s+10
(3) Completing the square yields Y=3
(s−1)2+9Thus
y(t) = e1tsin(3t)
Answer(s) submitted:
•
•
•
(incorrect)
Correct Answers:
•sˆ2*Y-3-2*s*Y+10*Y
•3/(sˆ2-2*s+10)
•exp(1*t)*sin(3*t)
8. (1 point) Use the Laplace transform to solve the following
initial value problem:
y00 +4y0+4y=0y(0) = 6,y0(0) = −1
(1) Using Yfor the Laplace transform of y(t), i.e.,
Y=L{y(t)}, find the equation you get by taking the
Laplace transform of the differential equation
=0
(2) Solve for
Y(s) =
(3) write the above answer in its partial fraction decompo-
sition, Y(s) = A
s+a+B
(s+a)2
Y(s) = +
(4) Now, by inverting the transform, find
y(t) = .
Solution:
(1) Taking the Laplace transform of both sides of the dif-
ferential equation yields:
s2Y−sy(0)−y0(0) + 4(sY −y(0)) + 4Y=0
Substituting the initial conditions gives:
s2Y−6s+4sY +4Y−23 =0
(2) Y=6s+23
s2+4s+4=6s+23
(s+2)2
(3) Setting up the partial fraction as
6s+23
(s+2)2=A
s+2+B
(s+2)2
yields A=6 and B=11.
(4) y(t) = 6e−2t+11te−2t
Answer(s) submitted:
•
•
•
•
•
(incorrect)
Correct Answers:
•sˆ2*Y-6*s--1+4*(s*Y-6)+4*Y
•(6*s+23)/(sˆ2+4*s+4)
•6/(s+2)
•11/(s+2)ˆ2
•6*exp(-2*t)+11*t*exp(-2*t)
3
9. (1 point)
Consider the initial value problem
y00 +16y=cos(4t),y(0) = 2,y0(0) = 4.
(1) Take the Laplace transform of both sides of the given
differential equation to create the corresponding alge-
braic equation. Denote the Laplace transform of y(t)
by Y(s). Do not move any terms from one side of the
equation to the other (until you get to part (b) below).
=help
(formulas)
(2) Solve your equation for Y(s).
Y(s) = L{y(t)}=
(3) Take the inverse Laplace transform of both sides of the
previous equation to solve for y(t).
y(t) =
Answer(s) submitted:
•
•
•
•
(incorrect)
Correct Answers:
•sˆ2*Y(s)-2*s-4+16*Y(s)
•s/(sˆ2+16)
•s/[(sˆ2+16)ˆ2]+(2*s+4)/(sˆ2+16)
•t/8*sin(4*t)+2*cos(4*t)+sin(4*t)
10. (1 point)
Use the Laplace transform to solve the following initial value
problem:
x0=6x+4y,y0=−2x+e4tx(0) = 0,y(0) = 0
Let X(s) = L{x(t)}, and Y(s) = L{y(t)}.
Find the expressions you obtain by taking the Laplace trans-
form of both differential equations and solving for Y(s)and
X(s):
X(s) =
Y(s) =
Find the partial fraction decomposition of X(s)and Y(s)and
their inverse Laplace transforms to find the solution of the sys-
tem of DEs:
x(t) =
y(t) = .
Solution: Taking the Laplace transform of both equations
and substituting the initial conditions, yields
sX =6X+4Y
sY =−2X+1
s−4
Solving for Yin the first equation gives Y=s−6
4X. Subsitut-
ing into the second equation yields
s(s−6)
4X+2X=1
s−4
s2−6s+8
4X=1
s−4
X=4
(s−4)(s2−6s+8)
X=4
(s−4)(s−4)(s−2)
X=4
(s−4)2(s−2)
Solving for Ygives
Y=s−6
(s−4)2(s−2)
Partial fraction decomposition yields
X=1
s−2+−1
s−4+2
(s−4)2
Y=−1
s−2+1
s−4+−1
(s−4)2
Thus the solution to the system is
x(t) = 1e2t−1e4t+2te4t
y(t) = −1e2t+1e4t−1te4t
Answer(s) submitted:
•
•
•
•
(incorrect)
Correct Answers:
•4/( (s-4)**2 * (s-2) )
•(s-6)/( (s-4)**2 *(s-2) )
•(4/4)*( exp(2*t)-exp(4*t)+ 2*t*exp(4*t) )
•(1/4)*( -4*exp(2*t)+4*exp(4*t)-2*2*t*exp(4*t) )
11. (1 point) Find the Laplace transform of te9tsin(8t).
L{te9tsin(8t)}=.
Solution: We will use the formula L{tnf(t)}=
(−1)nF(n)(s)with n=1.
Here f(t) = e9tsin(8t)and F(s) = L{f(t)}=8
(s−9)2+64
Differentiating F(s)yields F0(s) = −16(s−9)
((s−9)2+64)2
Thus
L{te9tsin(8t)}=−F0(s) = 16(s−9)
((s−9)2+64)2
Answer(s) submitted:
4
•
(incorrect)
Correct Answers:
•2*8*(s-9)/((s-9)**2+8**2)**2
12. (1 point) Find the Laplace transform of t2sin(4t).
L{t2sin(4t)}=.
Solution: We will use the formula L{tnf(t)}=
(−1)nF(n)(s)with n=2.
Here f(t) = sin(4t)and F(s) = L{f(t)}=4
s2+16
Differentiating F(s)yields F0(s) = −8s
(s2+16)2and F00(s) =
8(3s2−16)
(s2+16)3
Thus
L{t2sin(4t)}= (−1)2F00(s) = 8(3s2−16)
(s2+16)3
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•2*4*(3*sˆ2-4**2)/(sˆ2+4**2)**3
13. (1 point)
Given that
Lcos(7√t)
√πt=e−12.25/s
√s
find the Laplace transform of qt
πcos(7√t).
Lnqt
πcos(7√t)o=.
Solution: Let f(t) = cos(7√t)
√πt.
First note that rt
πcos(7√t) = t·f(t)
Thus, Lnqt
πcos(7√t)o=−F0(s)where F(s) = e−12.25/s
√s.
Thus
L(rt
πcos(7√t))=e−12.25/s2s−72
4s5/2
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•(2*s-49)*exp(-(12.25/s))/(4*sˆ2.5)
5
