# Transforms of Derivatives and Integrals. Differential Equations

```AB2.8: Transforms of Derivatives and Integrals. Differential
Equations
THEOREM 5.2.1
Suppose that f (t) is continuous for all t ≥ 0 satisfies the condition
|f (t)| ≤ M ekt
(t ≥ 0)
for some constants k and M . and has a derivative f 0 (t) that is piecewise continuous on any
finite interval in the range t ≥ 0. Then the Laplace transform of the derivative f 0 (t) exists
when s &gt; k, and
L (f 0 ) = sL (f ) − f (0),
s &gt; k.
PROOF
We first consider the case when f 0 (t) is continuous for all t ≥ 0. Then, by definitions and
integration by parts, we obtain the desired result
L (f 0 ) =
Z ∞
0
∞
e−st f 0 (t)dt == e−st f (t) + s
0
Z ∞
0
e−st f (t)dt = 0 − f (0) + sL (f ), s &gt; k.
By applying the theorem to the second derivative, we obtain
L (f 00 ) = sL (f 0 ) − f 0 (0) = s[sL (f ) − f (0)] − f 0 (0) = s2 L (f ) − sf (0) − f 0 (0).
Similarly,
L (f 000 ) = s3 L (f ) − s2 f (0) − sf 0 (0) − f 00 (0).
Using the mathematical induction, we obtain the Laplace transforms for the general case of
the derivatives f (n) of an arbitrary order n = 1, 2, . . .
L (f (n) ) = sn L (f ) − sn−1 f (0) − sn−2 f 0 (0) − . . . − sf (n−2) (0) − f (n−1) (0).
EXAMPLE 1 Laplace transform of f (t) = t2
Let f (t) = t2 , t ≥ 0. Find F (s).
Solution. We have
f 0 (t) = 2t,
f 00 (t) = 2,
so that
f (0) = 0,
f 0 (0) = 0,
f 00 (0) = 2.
Therefore,
L (f 00 ) = L (2) =
2
= s2 L (f ),
s
Resolving the latter equality with respect to s, we obtain the result
L (t2 ) =
2
.
s3
EXAMPLE 2 Laplace transform of cos ωt
Derive the Laplace transform of f (t) = cos ωt.
Solution. We have
f 00 (t) = −ω 2 cos ωt = −ω 2 f (t),
and
f (0) = 1,
f 0 (0) = 0.
Therefore,
L (f 00 ) = −ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ) − s.
Resolving the latter equality with respect to L, we obtain the result
L (f ) =
s
.
s2 + ω 2
EXAMPLE 2’ Laplace transform of sin ωt
Derive the Laplace transform of f (t) = sin ωt.
Solution. We have
f 0 (t) = ω cos ωt,
f 00 (t) = −ω 2 sin ωt = −ω 2 f (t),
so that
f (0) = 0,
f 0 (0) = ω.
Therefore,
L (f 00 ) = −ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ) − ω.
Resolving the latter equality with respect to L, we obtain the result
L (f ) =
s2
ω
.
+ ω2
EXAMPLE 3
Find the Laplace transform of f (t) = sin2 t.
Solution. We have
f 0 (t) = 2 sin t cos t = sin 2t,
f 00 (t) = 2 cos 2t,
so that
f (0) = 0,
f 0 (0) = 0,
f 00 (0) = 2.
Therefore,
L (f 00 ) = 2L (cos 2t) =
2s
= s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ).
+4
s2
Resolving the equality
2s
= s2 L (f )
+4
s2
with respect to s, we obtain the result
2
.
+ 4)
L (sin2 t) =
s(s2
EXAMPLE 4
Find the Laplace transform of f (t) = t sin ωt.
Solution. We have f (0) = 0 and
f 0 (t) = sin ωt + ωt cos ωt,
f 0 (0) = 0;
f 00 (t) = 2ω cos ωt − ω 2 t sin ωt = 2ω cos ωt − ω 2 f (t),
so that
L (f 00 ) = 2ωL (cos ωt) − ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ).
Resolving the equality
s
− ω 2 L (f ) = s2 L (f )
2
+ω
with respect to L, we obtain the result
2ω
s2
L (f ) = L (t sin ωt) =
(s2
2ωs
.
+ ω 2 )2
Differential Equations. Initial value problems
Begin with the initial value problem
y 00 + ay 0 + by = r(t),
y(0) = K0 ,
y 0 (0) = K1
(a, b = const).
(1)
Describe Laplace’s method for solving this nitial value problem.
First step. Writing
Y = L, R = L (r)
we apply the Laplace transform to (1) and obtain
L (y 00 ) + aL (y 0 ) + bL (y) = R
or
[s2 Y − sy(0) − y 0 (0)] + a[sY − y(0)] + bY = R.
This is called the subsidiary equation and reduces to
(s2 + as + b)Y = (s + a)y(0) + y 0 (0) + R(s).
Second step. Solve the subsidiary equation using the transfer function
Q = Q(s) =
s2
1
+ as + b
(2)
to obtain
Y (s) = [(s + a)y(0) + y 0 (0)]Q(s) + R(s)Q(s).
If y(0) = y 0 (0) = 0 this is simply Y = RQ and Q is the quotient
Q=
Y
L(output)
=
R
L(input)
Third step. Determine
y(t) = L−1 (Y )
by rearranging (3) algebraically to a sum of elementary transforms.
EXAMPLE 5
Solve the initial value problem
y 00 − y = t,
y(0) = 1,
y 0 (0) = 1.
Solution. In this case, we have, in terms of (1),
a = 0,
b = −1,
r(t) = t,
K0 = K1 = 1.
Apply Laplace’s method for solving this nitial value problem.
First step. Write
1
Y = L, R = L (r) = L (t) = 2
s
00
and apply the Laplace transform to y − y = t to obtain
L (y 00 ) − L (y) = R
and then the subsidiary equation
[s2 Y − sy(0) − y 0 (0)] − Y =
or
1
.
s2
1
.
s2
Second step. Solve the subsidiary equation using the transfer function
(s2 − 1)Y = s + 1 +
Q = Q(s) =
s2
1
−1
to obtain
Y (s) = [s + 1]Q(s) +
s+1
1
1
1
1
1
Q(s) = 2
+ 2 2
=
+ 2
− 2.
2
s
s − 1 s (s − 1)
s−1 s −1 s
Third step. Determine
−1
y(t) = L
−1
(Y ) = L
1
1
1
+ L−1 2
− L−1 2 = et + sinh t − t.
s−1
s −1
s
(3)
Laplace transform of the integral of a function
THEOREM 5.2.2
Let F (s) be the Laplace transform of f (t). If f (t) is piecewise continuous on any finite interval
in the range t ≥ 0 and satisfies the condition
|f (t)| ≤ M ekt
(t ≥ 0)
for some constants k and M , then
L
Z t
f (τ )dτ
0
or
Z t
1
= F (s) (s &gt; 0, s &gt; k),
s
f (τ )dτ = L
−1
0
1
F (s)
s
PROOF
If f (t) is piecewise continuous and satisfies the condition
|f (t)| ≤ M ekt
for k &gt; 0 and M &gt; 0, the integral
g(t) =
Z t
(t ≥ 0)
f (τ )dτ
0
is continuous and from the above inequality, we obtain
Z t
Z t
M kt
M kt
|g(t)| ≤
(e − 1) ≤
e
(k &gt; 0),
|f (τ )|dτ ≤ M
ekτ dτ =
k
k
0
0
which means that g(t) also satisfies this inequality. Also g 0 (t) = f (t); consequently, g 0 (t) is
piecewise continuous on each finite interval in the range t ≥ 0 and
L [f (t)] = L [g 0 (t)] = sL (g) − g(0)
(s &gt; k).
Here, g(0) = 0 and
F (s) = L (f ) = sL (g)
and finally
L (g) = L
Z t
f (τ )dτ
0
1
= F (s).
s
EXAMPLE 7
Let
L (f ) =
Find f (t).
Solution. We have
L
so that
L
−1
&quot;
−1
s(s2
1
.
+ ω2)
1
1
= sin ωt
2
2
s +ω
ω
#
1Zt
1 Zt
1
=
sin
ωτ
dτ
=
(1 − cos ωt).
s(s2 + ω 2 )
ω 0
ω2 0
PROBLEM 5.2.1
Solve the initial value problem
y 0 + 3y = 10 sin t,
y(0) = 0.
Solution. Apply Laplace’s method for solving this nitial value problem.
First step. Write
10
Y = L, R = L (10 sin t) = 2
s +1
0
and apply the Laplace transform to y + 3y = 10 sin t to obtain
L (y 0 ) + 3L (y) =
10
+1
s2
and then the subsidiary equation
[sY − y(0)] + 3Y =
or
10
.
+1
s2
10
.
+1
Second step. Solve the subsidiary equation using the transfer function
(s + 3)Y =
s2
Q = Q(s) =
to obtain
Y (s) =
1
s+3
1
10
10
A
Bs + C
Q(s) = 2
=
+ 2
.
+1
s +1s+3
s+3
s +1
s2
Find A, B, C by equating
10 = A(s2 + 1) + (Bs + C)s + 3,
which yields
A = 1,
B = −1,
C = 3.
Third step. Determine the desired solution
−1
y(t) = L
−1
(Y ) = L
1
s
1
− L−1 2
+ 3L−1 2
= e−3t − cos t + 3 sin t.
s+3
s +1
s +1
PROBLEM 5.2.5
Solve the initial value problem
y 00 + ay 0 − 2a2 y = 0,
y(0) = 6,
y 0 (0) = 0
(a = const).
Solution. In this case, we have, in terms of (1),
b = −2a2 ,
r(t) = 0,
K0 = 6,
K1 = 0.
Apply Laplace’s method for solving this nitial value problem.
First step. Write
Y = L(y),
R = L (r) = 0
and apply the Laplace transform to y 00 + ay 0 − 2a2 y = 0 to obtain
L (y 00 ) + aL (y 0 ) − 2a2 L (y) = 0
and then the subsidiary equation
[s2 Y − sy(0) − y 0 (0)] + a[sY − y(0)] − 2a2 Y = 0.
or
(s2 + as − 2a2 )Y = 6(s + a).
Second step. Solve the subsidiary equation using the transfer function
Q = Q(s) =
s2
1
+ as − 2a2
to obtain
Y (s) = 6(s + a)Q(s) = 6
Third step. Determine
y(t) = L−1 (Y ) = 6
6
s2
s2
s+a
+ as − 2a2
s
a
+6 2
=
2
+ as − 2a
s + as − 2a2
s
a
+6
=
2
2
2
(s +
− (a/2) − 2a
(s + a/2) − (a/2)2 − 2a2
s
a
6
+6
=
2
2
2
(s + a/2) − (3a/2)
(s + a/2) − (3a/2)2
a/2)2
6e−(a/2)t cosh (3a/2)t + 4e−(a/2)t sinh (3a/2)t = 2e−2at + 4eat .
PROBLEM 5.2.11
Find
F (s) = L (cos2 t).
Solution. Set
f = cos2 t,
cos2 t = 1 − sin2 t,
f (0) = 1,
f 0 = −2 cos t sin t = −2 sin 2t.
Now
2
= sL (f ) − 1.
+4
Resolving the latter equality with respect to L yields
L (− sin 2t) = −
s2
1
2
1
L (f ) =
1− 2
=
s
s +4
s
!
s2 + 2
.
s2 + 4
PROBLEM 5.2.12a
Find the Laplace transform of f (t) = t cos ωt.
Solution. We have f (0) = 0 and
f 0 (t) = cos ωt − ωt sin ωt,
f 0 (0) = 1;
f 00 (t) = −2ω sin ωt − ω 2 t cos ωt = −2ω sin ωt − ω 2 f (t),
so that
L (f 00 ) = −2ωL (sin ωt) − ω 2 L (f ) = s2 L (f ) − sf (0) − f 0 (0) = s2 L (f ) − 1.
Resolving the equality
2ω 2
− ω 2 L (f ) = s2 L (f ) − 1
2
2
s +ω
with respect to L, we obtain the result
−
L (f ) = L (t cos ωt) =
s2 − ω 2
.
(s2 + ω 2 )2
PROBLEM 5.2.12b
Prove that
−1
L
!
1
2
(s + ω 2 )2
Solution. We have
=
1
(sin ωt − ωt cos ωt).
2ω 3
ω(s2 − ω 2 )
,
(s2 + ω 2 )2
ω
,
L (sin ωt) = 2
s + ω2
−L (ωt cos ωt) = −
so that
ω
ω(s2 − ω 2 )
2ω 3
L (sin ωt − ωt cos ωt) = 2
− 2
= 2
,
s + ω2
(s + ω 2 )2
(s + ω 2 )2
which yields the desired formula.
PROBLEM 5.2.13
Find f (t) for
F (s) = L (f ) =
Solution. We have
L
Z t
and
f (t) =
−4τ
e
0
Z t
0
1
+ 4s
1 1
1
= L (e−4t ),
ss+4
s
F (s) =
so that
s2
dτ
1
= L (e−4t ),
s
1
e−4τ dτ = (1 − e−4t ).
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