Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 6.5 Impulse Function due 04/07/2021 at 11:59pm MST
1. (1 point)
Use the Laplace transform to solve the following initial value
problem:
y00 −4y0−21y=δ(t−7)y(0) = 0,y0(0) = 0
y(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution: Let Y=L{y(t)}. Taking the Laplace transform of
both sides of the differential equation yields:
s2Y−sy(0)−y0(0)−4(sY −y(0)) −21Y=e−7s
Substituting the initial conditions gives:
s2Y−4sY −21Y=e−7s
.
Thus
Y=e−7s1
s2−4s−21 =e−7s1
(s−7)(s+3)
Partial Fraction Decomposition gives
1
(s−7)(s+3)=
1
10
s−7+−1
10
s+3
Since
L−1(1
10
s−7+−1
10
s+3)=1
10 e7t−1
10 e−3t
,
it follows that
y(t) = L−1{Y}=L−1e−7s1
10
s−7+−1
10
s+3
=u(t−7)1
10 e7(t−7)−1
10 e−3(t−7)
Answer(s) submitted:
•u(t-7)((1/(10))eˆ(3(t-7))-(1/(10))eˆ(-7(t-7)))
(incorrect)
Correct Answers:
•u(t-7)*[0.1*exp(-(-7)*(t-7))+-0.1*exp(-3*(t-7))]
2. (1 point)
Use the Laplace transform to solve the following initial value
problem:
y00 +9y=7δ(t−4)y(0) = 0,y0(0) = 0
y(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution: Let Y=L{y(t)}. Taking the Laplace transform of
both sides of the differential equation yields:
s2Y−sy(0)−y0(0) + 9Y=7e−4s
Substituting the initial conditions gives:
s2Y+9Y=7e−4s
.
Thus
Y=e−4s7
s2+9=e−4s7
3
3
s2+9
Since
L−17
3
3
s+9=7
3sin(3t),
it follows that
y(t) = L−1{Y}=L−1e−4s7
3
3
s+9=7
3u(t−4)·sin(3(t−4))
Answer(s) submitted:
•u(t-4)*(7/3)*sin(3*(t-4))
(correct)
Correct Answers:
•u(t-4)*2.33333*sin(3*(t-4))
3. (1 point)
Use the Laplace transform to solve the following initial value
problem:
y00 +12y0+45y=δ(t−3)y(0) = 0,y0(0) = 0
y(t) = .
(Notation: write u(t-c) for the Heaviside step function uc(t)
with step at t=c.)
Solution: Let Y=L{y(t)}. Taking the Laplace transform of
both sides of the differential equation yields:
s2Y−sy(0)−y0(0) + 12(sY −y(0)) + 45Y=e−3s
Substituting the initial conditions gives:
s2Y+12sY +45Y=e−3s
.
Thus
Y=e−3s1
s2+12s+45
1
Completing the square gives
1
s2+12s+45 =1
(s+6)2+32
Since
L−11
(s+6)2+32=1
3
3
(s+6)2+32=1
3e−6tsin(3t),
it follows that
y(t) = L−1{Y}=L−1ne−3s1
3
3
(s+6)2+32o
=1
3u(t−3)e−6(t−3)sin(3(t−3))
Answer(s) submitted:
•(1/6)u(t-3)eˆ(-5(t-3))sin(6(t-3))
(incorrect)
Correct Answers:
•u(t-3)*0.333333*exp(-6*(t-3))*sin(3*(t-3))
2
