2009/10 Schol paper 1, my questions, outline
solutions.
1. The Laplace transform of a function f (t), t ≥ t is a function of s defined
as
Z ∞
f (t)e−st dt
L[f (t)] =
0
Show
1
s−a
where a is a constant and you may assume s > a. Show that,
L[eat ] =
#
"
df
= sL[f (t)] − f (0)
L
dt
where you can assume
lim f (t)e−st = 0
t→∞
By taking the Laplace transform of both sides of the equation solve
df
= 2f
dt
where f (0) = −1.
Solution:So you just have to follow this question along; first
Z
L[eat ] =
∞
0
e(a−s)t dt =
i∞
1
1
e(a−s)t =
0
a−s
s−a
(1)
where we have assumed that s > a in order to evaluate the exponential
at infinity.
Next, from the definition
"
#
df
=
L
dt
Now use integration by parts
"
df
L
dt
#
=
Z
0
∞
df −st
e dt
dt
1
Z
∞
0
df st
e dt
dt
(2)
= f (t)e−st
i∞
0
−s
Z
0
∞
df −st
e dt = sL[f (t)] − f (0)
dt
(3)
Finally taking the Laplace transform of both sides of the differential
equation gives
sL[f (t)] + 1 = 2L[f (t)]
(4)
so, solving gives
1
(5)
s−2
and, hence, by the above, and noting that the transform is linear
L[f (t)] + 1 = −
f (t) = −e2t
(6)
2. Write the on-off pulse



as a Fourier integral.
1
−π < t < 0
f (t) = −1 0 < t < π


0
otherwise
Solution:Well lets apply the definition and integrate
Z
1 ∞
f (t)e−ikt dt
2π Z−∞
1 0 −ikt
1 Z π −ikt
=
e dt −
e dt
2π −π
2π 0
0
1 1 −ikt
1 1 −ikt π
=
−
e
e
2π −ik
2π −ik
−π
0
i i
=
1 − eikπ −
e−ikπ − 1
2πk
2πk
fg
(k) =
(7)
Now, putting all that together
fg
(k) =
i
i
−
cos πk
πk πk
3. Find the general solution of
d2 y
dy
d3 y
−
3
+ 3 − y = et
3
2
dt
dt
dt
2
(8)
Solution:First off, substitute into the homogenous equation
y = eλt
(9)
λ3 − 3λ2 + 3λ − 1 = 0
(10)
(λ − 1)3 = 0
(11)
giving
which factorizes as
so the homogenous equation has general solution
y = (C1 + tC2 + t2 C3 )et
(12)
Hence, to match the right hand side we need to substitute
y = Ct3 et
(13)
y ′ = Ct3 et + 3Ct2 et
y ′′ = Ct3 et + 6Ct2 et + 6Ctet
y ′′′ = Ct3 et + 9Ct2 et + 18Ctet + 6Cet
(14)
so
All the terms with t outside the exponential in them cancel and we get
6C = 1
(15)
and hence the solution to the inhomogenous equation is
1
y = C1 + tC2 + t2 C3 + t3 et
6
(16)
4. Find the general solution of
t2 ÿ + 3tẏ + y = t2
Solution:So this is an Euler equation, let
so
t = ez
(17)
dy dz
1 dy
dy
=
=
dt
dz dt
t dz
(18)
3
and
d2 y
d
=
2
dt
dt
1 dy
t dz
!
=−
1 dy
1 d2 y
+
t2 dz t2 dz 2
(19)
Now, the equation becomes
d2 y
dy
+ 2 + y = e2z
2
dz
dz
(20)
which can be solved using the normal substitution to give
or
1
y = (C1 + C2 z)e−z + e2z
9
(21)
1 1
y = (C1 + C2 log t) + t2
t 9
(22)
4