Infinite series and differential equations

g. 
Therefore, if two continuous functions have the same Laplace transforms, they are completely 
identical. 
This means that, omitted the discontinuous points of the functions, we have that the relation 
between an original function and its Laplace transform is one-to-one. 
Thus, the original function f(t) in (1) is called the inverse Laplace transform of F and is 
denoted by L -1(F). It is proved that, under some conditions, the original function f(t) can be 
reconstructed from F(s) by the formula 





i
i
stdsesF
i
tf )(
2
1)( for some large enough . 
We note that the original functions are denoted by lowercase letters, and their Laplace 
transforms--by the same letters in capitals, e.g., F= L(f), G= L(g), etc. 
2. Properties 
2.1. Linearity: For all piecewise continuous functions f, g, and constants a, b we have 
L(af+bg)= aL(f)+bL(g). 
Physically, one writes: af(t)+bg(t) aF(s)+bG(s). 
Examples: 1) Let f(t)=cosh(at)=(eat+ e-at)/2. Find F= L(f). 
We already have eat
as 
1
 and e-at
as 
1
. Therefore, 
2
1
(eat+ e-at ) 
2
1
(
as 
1
+
as 
1
). 
Hence, F(s)= 
2
1
(
as 
1
+
as 
1
)= 22 as
s
 . 
2) Let F(s)= 
))((
1
bsas  ; a  b. Find f=L 
-1(F). 
We first write F(s)= 
))((
1
bsas  = )
11(
)(
1
bsasba  . Therefore, by the linearity, we 
obtain 
ba 
1
(eat-ebt ) 
ba 
1
(
as 
1
-
bs 
1
). Thus, f(t)= 
ba 
1
(eat-ebt ). 
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Nguyen Thieu Huy 
2.2. Laplace transform of the derivative of f(t): 
Theorem 1: Let f be differentiable and exponentially bounded. 
 If f(t) F(s), then f´(t) sF(s)-f(0) for s>0. 
Proof: (L f´)(s) =     
0 0
0
)(f)('f dtetsedtet ststst =sF(s)-f(0). (qed) 
If the second derivative f´´exists, applying the above formula for f´we obtain that 
f´´(t) s2F(s)-sf(0)-f´(0). 
Generally, by induction we have the following theorem. 
Theorem 2: In case the nth derivative of f exists and exponentially bounded, we obtain the 
formula 
f(n)(t) snF(s)-sn-1f(0)-sn-2f´(0)--f(n-1)(0). 
Examples: 1) For f(t)=t2 we find F= L f. To do so, we observe that f´(t)=2t; f´´(t)=2. 
We already have 2=f´´(t) 2/s. Therefore, 2/s = s2F(s)-sf(0)-f´(0). It follows that 
F(s)=2/s3. 
2) Similarly, we easily obtain that sinwt 22 ws
w
 . 
2.3. Laplace transform of the integral of f(t): 
Theorem 3: If f(t) F(s), then t duu
0
)(f
s
1 F(s). 
Proof: Put g(t)= . Then, g´(t)=f(t) and g(0)=0. Let F = L f and G= L g. By 
Theorem 1, we obtain that g´(t)
t duu
0
)(f
 sG(s)-g(0)=F(s). Therefore, G(s)=F(s)/s. (qed) 
Example: For F(s) = 
)(
1
22 wss  let find f(t). We already have w
wtsin
22
1
ws  . 
By Theorem 3, we then derive t duwwu0
sin
)(
1
22 wss  . Thus, 
2
cos1
w
wt
)(
1
22 wss  . 
2.4. Inverse Laplace transform of the derivative of F(s): 
For f(t) F(s) we can easily prove that – tf(t) F´(s). 
Example: Since sinwt 22 ws
w
 we have, by the above formula, that 
tsinwt 222 )(
2
ws
sw
 . 
2.5. Inverse Laplace transform of the integral of F(s): 
For f(t) F(s) it can be proved that 
t
tf )( 

s
duuF )( . 
 73
 Lecture on Infinite Series and Differential Equations 
Example: Let compute the inverse Laplace transform of G(s) = ln(1+ )2
2
s
w
. To do that, we 
first write ln(1+ )2
2
s
w
=- 

 ss
du
wuu
w
u
wd
)(
2)1ln( 22
2
2
2
. Since 2
cos1
w
wt
)(
1
22 wss  
and applying the above formula we obtain 
t
wt)cos1(2 
ln(1+ )2
2
s
w
. 
2.6. Shifting properties: 
(1) s-shifting: For f(t) F(s) we have that tse 0 f(t) F(s-s0) 
Example: From sinwt 22 ws
w
 we obtain that e sinwt
ts0
22
0 )( wss
w
 
(2) t-shifting: If we shift the function f(t), t0, to the right (i.e., we replace t by t-a for 
some a>0), then we encounter a problem that the function f(t-a) is no longer defined 
for a>t0. To come over this problem, we put . 



atifatf
atif
tf
)(
00
)(
~
using the step function u(t)= we can rewrite =f(t-a)u(t-a). Then, for 
f(t)




01
00
tif
tif
)(tf
~
F(s) it can be proved that f(t-a)u(t-a) sae F(s). 
Example: Let compute the inverse Laplace transform of 3
2
s
e s
. From the relation 
t2/2 1/s3 we can derive that 
2
)2( 2t
u(t-2) 3
2
s
e s
. 
3. Convolution 
3.1. Lemma: Let f(t) and g(t), t0, be piecewise continuous and exponentially bounded 
functions. Then, the function h(t) = is also exponentially bounded.  t duutguf
0
)()(
PROOF: Since f(t) and g(t) are piecewise continuous and exponentially bounded, we can 
estimate |h(t)| = t utu dueeM
0
)(
1
21  ||
||
21
21
21 

tt eeMM  =
},max{
21
21 21
||
2 

teMM . Therefore, h(t) is 
exponentially bounded. 
3.2. Definition: Let f(t) and g(t), t0, be piecewise continuous and exponentially bounded 
functions. Then the function h(t) = is called convolution of f and g. Also, we 
denote by h=fg. So, h(t)=(fg)(t). However, sometimes, physically we write h(t)=f(t)g(t). 
 t duutguf
0
)()(
3.3. Theorem: Let f(t) and g(t), t0, be piecewise continuous and exponentially bounded 
functions. Suppose F = L f and G= L g. Then, f(t)g(t) F(s) G(s). 
Shortly, one can say that Laplace transform turns a convolution to a normal product. 
  
 




 
00 00
)()()()())((
u
st
t
stst dtdueutgufdudteutgufdtetgf PROOF: Let compute 
 74
Nguyen Thieu Huy 
 = F(s)G(s) 
here, we used Fubini’s Theorem for the domain described by following figure: 
= )()(
0
sGdueuf su 


  
. Since we already have 
w
wtsin
22
1
ws Example: Let compute L 
-1 


 222 )(
1
ws
, 
ty we obtainusing the convolution proper 
w
wtsin
w
wtsin
22 )
1
w
. 2(s 
3.4. Some other properties: 
k 
ential Equations 
e have the following algorithm of using Laplace transform to solve differential equations of 
ry equation. 
tep 2: Solve the subsidiary equation. 
l 
aplace transform to the above equation and putting Y = Ly we obtain 
2 2
1) Associative: (fg)k = f(gk) 
2) Commutative: fg = gf 
3) Distributive: f(g + k) = g f + f
4. Applications to Differ
W
the order n: f(t, y, , y(n))=r(t). 
Step 1: Apply the Laplace transform to both sides of the differential equation to obtain the 
simpler equation called subsidia
S
Step 3: Apply the inverse Laplace transform to obtain the solution of the original differentia
equation. 
Example 1: y’’-y=t; with the initial conditions y(0)=1; y’(0)=1; 
Applying L
s Y(s)-sY(0)-y’(0)-Y(s)=1/s 
Y(s)= 2222 11
1111
s
 1)1(1 sssss 
btain that Using the table of Laplace transforms we easily o
y(t)= L-1Y=et + sinht-t. 
aplace transform, we now 
sing Laplace transform: 
Before continuing with further examples of applications of L
introduce here a scheme for solving a differential equation u
 75
 Lecture on Infinite Series and Differential Equations 
Example 2: Consider the general linear second-order differential equation with constant 
coefficients: 
y’’ + ay’ + by=r(t); with initial conditions y(0) = y0, y’(0) = y1. (4.1) 
Applying Laplace transform to both sides of the given equation and putting Y= L y, R= L r, 
we obtain the subsidiary equation: (s2 + as + b)Y(s)=R(s) + (s+a)y0 + y1. Therefore, we have 
that the solution of subsidiary equation is 
Y(s) = 
b as s
y a)y(s R(s)
2
10


. 
 Therefore, we obtain the solution of the given differential equation by taking the inverse 
Laplace transform of Y(s), i.e., the solution is y= L -1Y. 
We now put Q(s) = 
b as s
1
2  and call it the transfer function. This name comes from the 
fact that, for some (mechanic or electric) systems, the function r(t) in equation (4.1) is called 
the input and the solution y(t) is called the output of the system, and in the special case when 
y(0)=0 and y’(0)=0, then Y(s)=R(s)Q(s). Therefore, Q(s)= L (output)/ L (input) explaining 
the name of Q(s). Also, in this case, the output is y(t)=r(t)q(t), where q(t) is inverse Laplace 
transform of Q(s). 
 76
Nguyen Thieu Huy 
Tables of Laplace Transform: 
Table 1: General Formulae 
 77
 Lecture on Infinite Series and Differential Equations 
Table 2: Laplace Transform 
 78
Nguyen Thieu Huy 
Table 3: Laplace Transform (continued) 
 79
 Lecture on Infinite Series and Differential Equations 
Problems 
In each of Problems 1 through 10 find the inverse Laplace transform of the given function. 
In each of Problems 11 through 23 use the Laplace transform to solve the given initial value 
problem. 
 80
Nguyen Thieu Huy 
In each of Problems 24 through 36 find the solution of the given initial value problem. 
26. y’’+ 4y = sin t − u(t-2π) sin(t − 2π); y(0) = 0, y’(0) = 0 
27. y’’+ 4y = sin t + u(t-π) sin(t − π); y(0) = 0, y’(0) = 0 
28. y’’ + 3y’ + 2y = f (t); y(0) = 0, y’(0) = 0; f (t) = 1 for 0 ≤ t < 10 and f (t) =0 for t ≥ 10 
29. y’’ + 3y’ + 2y = u(t-2); y(0) = 0, y’(0) = 1 
30. y’’ + y = u(t-3π); y(0) = 1, y’(0) = 0 
31. y’’ + y’ + 
4
5 y = t − u(t - π/2)(t − π/2); y(0) = 0, y’(0) = 0 
32. y’’ + y = g(t); y(0) = 0, y’(0) = 1; g(t) = t/2 for 0 ≤ t < 6 and g(t) = 3 for t ≥ 6, 
33. y’’ + y’ + 
4
5 y = g(t); y(0) = 0, y’(0) = 0; g(t) = sin t for 0 ≤ t < π and g(t) = 0 for t ≥ π 
34. y’’ + 4y = u(t-π) − u(t-3π); y(0) = 0, y’(0) = 0 
35. y(4) − y = u(t-1) − u(t-2); y(0) = 0, y’(0) = 0, y’’(0) = 0, y’’’(0) = 0 
36. y(4) + 5y’’ + 4y = 1 − u(t-π); y(0) = 0, y’(0) = 0, y’’(0) = 0, y’’’(0) = 0 
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 Lecture on Infinite Series and Differential Equations 
 82
References 
1. W. E. Boyce, R. C. DiPrima: Elementary Differential Equations and Boundary Value 
Problems, 7th ed., John Wiley & Sons, 2001. 
2. R. Bronson: Differential Equations, The McGraw-Hill, 2003. 
3. P. O’Neil: Advanced Engineering Mathematics, 5th Edition, Thomson, 2003. 
4. R. Wrede, M. R. Spiegel: Theory and Problems of ADVANCED CALCULUS, 2nd Edition, 
McGraw-Hill, 2002.