MATH2019 Problem 7 Solution

EXAMPLES7
LAPLACETRANSFORMS
2014, S1 1. Using the table of Laplace transforms find
.
2014, S1
2014, S2
2014, S2
2015, S1
2015, S1
2015, S1
2. i) By establishing an appropriate partial fraction decomposition find
.
ii) Hence or otherwise find
.
3. Find:
i) L{et cos(πt) + et sin(πt)}.
ii) .
4. Use the Laplace transform method to solve the initial value problem
y′′ − y′ = 4u(t − 2) with y(0) = 1, y′(0) = 1,
where u(t − 2) is a Heaviside step function.
5. Find:
i) L{t3eπt}.
ii) .
6. The function f(t) is given by
0 for 0 ≤ t < 1, f(t) =t − 1 for 1 ≤ t < 3,  2 for t ≥ 3.
i) Sketch the function f(t) for 0 ≤ t ≤ 4.
ii) Write f(t) in terms of the Heaviside step function u(t − a). iii) Hence, or otherwise, find the Laplace transform of f(t).
7. Use the Laplace transform method to solve the initial value problem
y′′ − 4y = 8u(t − 1) with y(0) = 1, y′(0) = 2,
where u(t − 1) is a Heaviside step function.
2015, S2 8. Find:
i) L{t5e3t}.
ii) .
2015, S2 9. The function g(t) is given by
t for 0 ≤ t < 1 e for t ≥ 1.
g(t) = ( t
i) Sketch the function g(t) for 0 ≤ t ≤ 2.
ii) Write g(t) in terms of the Heaviside step function. iii) Hence, or otherwise, find the Laplace transform of g(t).
2015, S2 10. i) Find the partial fraction decomposition of
30
.
s(s + 3)(s − 2)
ii) Using the Laplace transform method and your answer in the previous part find thesolution of the initial value problem
4) with y(0) = 0 and y′(0) = 0,
where u(t − 4) is the Heaviside step function.
∞ F(s) = L{f(t)} =f(t)e−stdt.
0
i) Use Leibniz’ rule to prove
L{tf(t)} = −F ′(s).
ii) Hence, or otherwise, find the following Laplace transform
L{tsin(3t)} .
y′′ − y = u(t − 1) with y(0) = 0, y′(0) = 1,
where u(t − 1) is a Heaviside step function.
i) L{e−3t sin(πt)}.
ii) .
14. The function g(t) is given by
for 0 ≤ t < 2,
g( ) =
−1 for t ≥ 2.
i) Sketch the function g(t) for 0 ≤ t ≤ 6.
ii) Write g(t) in terms of the Heaviside step function u(t − a).
iii) Hence, or otherwise, show that the Laplace transform of g(t) is given by
.
The vertical velocity v(t) of the rocket satisfies the differential equation

where g(t) is the function in the previous question.
i) Using Laplace transforms and the result from part iii) in the previous question, solvethe differential equation above to find the velocity of the rocket v(t) as a function of time.
ii) By writing your solution separately for times 0 ≤ t < 2 and t ≥ 2, or otherwise, sketch the velocity as a function of time for 0 ≤ t ≤ 6.
iii) What is the maximum velocity of the rocket?iv) At what time will the rocket reach its maximum height?
i) L{t u(t − 2)}.
ii) .
1 for 0 ≤ t < 1,
g( ) = −t+1 e for t ≥ 1.
i) Sketch the function g(t) for 0 ≤ t ≤ 3.
ii) Write g(t) in terms of the Heaviside step function u(t − a). iii) Hence, or otherwise, show that the Laplace transform of g(t) is
.
y′′ − y′ = g(t), y(0) = −1, y′(0) = 0,
where g(t) is the function from the previous question.

19. i) The Laplace transform of a function f(t) is defined for t ≥ 0 by

Prove directly from the above definition that

where a > 0 and u(t − a) is the Heaviside function. ii) Find:
.
) for 0 ≤ t < 1,
g( ) =
0 for t ≥ 1.
i) Sketch the function g(t) for 0 ≤ t ≤ 2.
ii) Write g(t) in terms of the Heaviside step function u(t − a). iii) Hence, or otherwise, show that the Laplace transform of g(t) is
.
[Hint: You can use sin(A + π) = −sinA.]
y′′ − y′ − 2y = 6u(t − 1), y(0) = 1, y′(0) = 2.
i) L{te−t sin(3t)};
ii) .
,
i) Express f(t) in terms of the Heaviside function.
y′′ − 4y′ + 4y = f(t), t > 0,
subject to the initial conditions y(0) = 1 and y′(0) = 0, where f(t) is given in the previous question.
25. Find
i) L{sin(3t)}, ii) L{e−7t sin(3t)},
iii) .
,
i) Express g(t) in terms of the Heaviside function. ii) Hence, or otherwise, show that the Laplace transform of g(t) is
.
TABLE OF LAPLACE TRANSFORMS AND THEOREMS
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TABLE OF LAPLACE TRANSFORMS AND THEOREMS
g(t) is a function defined for all t ≥ 0, and whose Laplace transform
∞
G(s) = L{g(t)} =e−stg(t)dt
0
exists. The Heaviside step function u is defined to be
0 for t < a
u(t − a) = (
1 for t > a
g(t) G(s) = L{g(t)}
1
t
1

s2
tm,m = 0,1,... m!

sm+1
e−αt 1

s + α
sin(ωt) ω

s2 + ω2
cos(ωt) s

s2 + ω2
u(t − a) e−as

s
f′(t) sF(s) − f(0)
f′′(t) s2F(s) − sf(0) − f′(0)
e−αtf(t) F(s + α)
f(t − a)u(t − a) e−asF(s)
tf(t) −F ′(s)