Many physical problems when analyzed assumes the form of an ordinary differential equation subjected to set of initial conditions or boundary conditions. Such problems are referred to as initial value problems and boundary value problems respectively. Laplace Transforms serves as a very useful tool in solving these problems without actually finding the general solution of the differential equation by various known methods. The application of Laplace Transforms is highly significant in engineering problems associated with electric circuit.

In this section let us study about the Definition of Laplace Transforms, Properties, Laplace Transforms of Derivatives, Integrals, Steps functions, constant functions and few solved examples.

## Laplace Transforms

Laplace Transforms are useful to solve the ordinary differential equations. When we encounter models for the dynamics which depend on rate of change of functions like velocities and accelerations of particles, which prompts the use of ordinary differential equations ( ODE ).Using Laplace transforms we convert the whole ODE initial value problems into a function by using the basic transformations. After the necessary algebraic steps we convert or use inverse Laplace Transform to find the required function f ( t ).

## Definition of Laplace Transform

If f ( t ) is a real valued function defined for all t $\ge$ 0 then the Laplace transform of f ( t ) is denoted by L[f(t)] is defined by

L [ f ( t ) ] = $\int_{t=0}^{\infty }$ $e^{-st}$ f ( t ) dt , provided the integral exists.
On integration of the indefinite integral we will be having a function of s and t. When this is evaluated between the limites t = 0 and t = $\infty$, we will be left with a function of s only and we shall denote by $\overline{f}$ (s), where s is a parameter, real or complex.
Thus we have L [ f ( t ) ] = $\overline{f}$ ( s )

## Laplace Transforms Properties

Since the linearity property holds good for an integral, the following properties hold good for Laplace Transformation also.

Therefore we have the following properties of Laplace Transformations.
1. L [ f1 ( t ) + f2 ( t ) ] = L [ f1 ( t ) ] + L [ f2 ( t ) ]

2. L [ f1 ( t ) - f2 ( t ) ] = L [ f1 ( t ) ] - L [ f2 ( t ) ]

3. L [ c f ( t ) ] = c L [ f ( t ) ]

## Laplace Transform Table

The following table shows the Laplace Transformation of different functions.
 f(t) a eat cos(at) sin(at) cosh(at) sinh(at) tn (for complex number) tn (for integer)( n = 1, 2, 3, . . . ) L[f(t)] $\frac{a}{s}$ $\frac{1}{s-a}$ $\frac{s}{s^{2}+a^{2}}$ $\frac{a}{s^{2}+a^{2}}$ $\frac{s}{s^{2}-a^{2}}$ $\frac{a}{s^{2}-a^{2}}$ $\frac{\Gamma (n+1)}{s^{n+1}}$ $\frac{n!}{s^{n+1}}$

The following examples will help us to transform into Laplace Transformations.
1. L ( 4 ) = $\frac{4}{s}$

2. L [ e2t ] = $\frac{1}{s-2}$

3. L ( e-t ) = $\frac{1}{s+1}$

4. L ( cos h 2t ) = $\frac{s}{s^{2}-4}$

5. L ( sin h 3t ) = $\frac{6}{s^{2}-4}$

6. L (t3 ) = $\frac{5!}{s^{6}}$

7. L ( 3 cos 4t ) = $\frac{3s}{s^{2}+16}$

8. L( sin 5t ) = $\frac{5}{s^{2}+25}$

9. L ($\sqrt{t}$ ) = $\frac{\Gamma (3/2)}{s^{3/2}}$ = $\frac{1/2.\Gamma 1/2)}{s^{3/2}}$ = $\frac{\sqrt{\pi}}{2.s^{3/2}}$.

## Inverse Laplace Transformation

We have already discussed that the Laplace Transform of f ( t ) is $\overline{f}$ ( s ), which is denoted as L [ f ( t ) ] = $\overline{f}$ ( s ).

Here f ( t ) is called the inverse Laplace Transform of $\overline{f}$ (s) and is denoted by L$^{-1}$ [ $\overline{f}$ ( s ) ].

In general, L [ f ( t ) ] = $\overline{f}$ (s) <=> L$^{-1}$ [ $\overline{f}$ (s) ] = f ( t )

The following examples help us how the Inverse Laplace Transformations give us the original function in terms of t.

1. L ( 1 ) = $\frac{1}{s}$

=> L-1 ( $\frac{1}{s}$ ) = 1

2. L (cos at ) = $\frac{s}{s^{2}+a^{2}}$ )

=> $L^{-1}$ $(\frac{s}{s^{2}+a^{2}})$ = cos at

## Laplace Transform of Derivatives

If L [ f ( t ) ] = $\overline{f}$ ( s ), then L [ tn . f ( t ) ] = ( -1 )n . $\frac{\mathrm{d^{n}} }{\mathrm{d} x^{n}}$ [ $\overline{f}$ ( s ) ], where n is a positive integer.

This property is called the derivative of transform property.

( i.e ) L [ t . f ( t ) ] = ( -1 )1 . $\frac{\mathrm{d} }{\mathrm{d} x}$ [ $\overline{f}$ ( s ) ]

L [ t2 . f ( t ) ] = ( -1 )2 . $\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}}$ [ $\overline{f}$ ( s ) ]
and so on.

### Solved Example

Question: Find the Laplace Transform of the function t cos(at).
Solution:

We have to find L [ t cos at ]
Let f ( t )  = cos at

Therefore,           L [ f ( t ) ] = $\frac{s}{s^{2}+a^{2}}$  = $\overline{f}$ ( s )

Now L [ t f ( t ) ] = - $\frac{\mathrm{d} }{\mathrm{d} x}$ [ $\overline{f}$ ( s ) ]

= - $\frac{\mathrm{d} }{\mathrm{d} x}$  $\frac{s}{s^{2}+a^{2}}$

= - [  $\frac{(s^{2}+a^{2})-s.2s}{(s^{2}+a^{2})^2}$ ]

Thus L [ t . cos at ]    = $\frac{s^{2}-a^{2}}{(s^{2}+a^{2})^{2}}$

## Laplace Transform of Integral

1. If L [ f ( t ) ] = $\overline{f}$ ( s ) , then L [ $\frac{f(t)}{t}$ ] = $\int_{s}^{\infty}$ $\overline{f}$ ( s ) ds

2. If L [ f ( t ) ] = $\overline{f}$ ( s ) then L [ $\int_{0}^{t}$ f ( t ) dt ] = $\frac{\overline{f}(s)}{s}$

### Solved Examples

Question 1: Show that $\int_{s}^{\infty}$ t . e-2t . sin4t dt = $\frac{1}{25}$
Solution:

We have $\int_{s}^{\infty}$ t  e-2t sin4t dt   = L ( t . sin 4t )

L ( t sin 4t ) = $\frac{-d}{ds}$ ($\frac{4}{s^{2}+16}$) = $\frac{8s}{(s^{2}+16)^{2}}$

Therefore,               $\int_{0}^{\infty}$ t  e-2t sin 4t dt = $\frac{8s}{(s^{2}+16)^{2}}$
Substituting s = 2, we get,
$\int_{0}^{\infty}$ t  e-2t sin 4t dt  = $\frac{16}{400}$ = $\frac{1}{25}$

Question 2: Show that L [ $\frac{sinh\;t}{t}$ ] = $\frac{1}{2}$ log $\frac{(s+1)}{(s-1)}$
Solution:

Let f ( t ) = sinh t

Therefore,                      f ( s ) = $\frac{1}{s^{2}-1}$

Hence L [ $\frac{sinh\;t}{t}$ ]    = $\int_{s}^{\infty}$ $\frac{1}{s^{2}-1}$ dx

= $\frac{1}{2}$ [ log ( $\frac{(s-1)}{(s+1)}$  ]$_{0}^{\infty }$

= $\lim_{s\to\infty }$ $\frac{1}{2}$ [ log  $\frac{s(1-1/s)}{s(1+1/s)}$  ] - [ log ( $\frac{(s-1)}{(s+1)}$  ]

= $\frac{1}{2}$ [ log 1 -  log ( $\frac{(s-1)}{(s+1)}$  ]

= $\frac{1}{2}$ $\frac{(s+1)}{(s-1)}$

= log $\sqrt{\frac{(s+1)}{(s-1)}}$

## Laplace Transform of Step Function

Unit Step Function (Heaviside function): The unit step function u (t - a) or Heaviside function H(t-a) is defined as follows.
u ( t - a ) = 0, t $\le$ a
= 1, t > a, where a is a positive constant.

Properties associated with the unit step function:
1. L [ u ( t - a ) ] = $\frac{e^{-as}}{s}$

2. L [ f ( t - a ) . u ( t - a ) ] = e-as $\overline{f}$ ( s ), where L [ f ( t ) ] = f ( s ).

Working Procedure for Problems:

### Type 1: To find L [F(t) u ( t - a ) ] where F ( t ) is a polynomial in t.

a. Let F ( t ) = f ( t - a ) which implies that F ( t + a ) = f ( t )
b. Replace t by t + a to obtain f ( t )
c. Find L [ f ( t ) ] = $\overline{f}$ ( s )
d. Therefore, we have L [ F ( t ) u ( t - a ) ] = e-as . $\overline{f}$ ( s )

### Type 2: Given that f ( t ) as a discontinuous function, to find L [ f ( t ) ] by expressing f ( t ) in terms of unit step function.

a. We express f ( t ) in terms of unit step function at different intervals
b. Then we find L [ f ( t ) ] as in Type 1, discussed above.

### Solved Example

Question: Find the Laplace transform of the function (4t + 5) . u (t - 3)
Solution:

Let f ( t - 3 ) = 4 t + 5
=> f ( t ) = 4 ( t + 3 ) + 5 = 4 t + 12 + 5 = 4 t + 17

Therefore, $\overline{f}$ ( s ) = $\frac{4}{s^{2}}$ + $\frac{17}{s}$

We have L [ f ( t - 3 ) u ( t - 3 ) ] = e -3s $\overline{f}$ {s} ; ( since a = 3 )

Therefore, L [ ( 4t + 5 ) . u ( t - 3 ) ] = e- as [ $\frac{4}{s^{2}}$ + $\frac{17}{s}$ ].

## Laplace Transform of a Constant

From the definition of Laplace transformation,
L[f(t)] = $\int_{t=0}^{\infty }$ $e^{-st}$ f ( t ) dt
To find the Laplace transformation of a constant ( say a ), let us assume f ( t ) = a

Therefore, L [ a ] = $\int_{t=0}^{\infty }$ $e^{-st}$ a dt

= a [ $\frac{e^{-st}}{-s}$ ]$_{0}^{\infty}$

= $\frac{-a}{s}$ ( 0 - 1 )

L [ a ] = $\frac{a}{s}$

Example: L ( 5 ) = $\frac{5}{s}$

L ( 1 ) = $\frac{1}{s}$

## Laplace Transformation Examples

### Solved Examples

Question 1: Find the Laplace Transform of e- 2t ( 2 cos 5t - sin 5t )
Solution:

Let f ( t ) = 2 cos 5t - sin 5t
L [ f ( t ) ] = L [ 2 cos 5t - sin 5t ]

= 2 L [cos 5t ] - L [ sin 5t ]

= 2 $\frac{s}{s^{2}+5^{2}}$ - $\frac{5}{s^{2}+5^{2}}$

= $\frac{2s-5}{s^{2}+25}$

L [ e- 2t  f ( t ) ] = $\frac{2s-5}{s^{2}+25}$ ]$_{s->s+2}^{}$

= $\frac{2(s+2)-5}{(s+2)^{2}+25}$

= $\frac{(2s-1)}{s^{2}+4s+29}$

Question 2: Evaluate $\int_{0}^{\infty}$ $\frac{cos\;6t-cos\;4t}{t}$ dt
Solution:

$\int_{0}^{\infty}$ $\frac{cos\;6t-cos\;4t}{t}$ dt = L [ $\frac{cos6t-cos4t}{t}$

L [ $\frac{cos6t-cos4t}{t}$ = $\int_{s}^{\infty}$ { $\frac{s}{s^{2}+6^{2}}$ - $\frac{s}{s^{2}+4^{2}}$

= $\frac{1}{2}$ [ log ( s2 + 36 ) - log (s2 + 16 ) ]$_{0}^{\infty}$

= $\frac{1}{2}$  [ log ( $\frac{s^{2}+36}{s^{2}+16}$ ) ]$_{0}^{\infty}$

= $\frac{1}{2}$ $\lim_{s->\infty}$ log ( $\frac{s^{2}\left ( 1+\frac{36}{s^{2}} \right )}{s^{2}\left ( 1+\frac{16}{s^{2}} \right )}$ - log ( $\frac{s^{2}+36}{s^{2}+16})$

= $\frac{1}{2}$ [ log 1 - log  ( $\frac{s^{2}+36}{s^{2}+16}$ ) ]

= $\frac{1}{2}$ log ( $\frac{s^{2}+16}{s^{2}+36}$ )

= log $\sqrt{\frac{s^{2}+16}{s^{2}+36}}$ )
Substituting s = 0, we get,

$\int_{0}^{\infty}$ $\frac{cos\;6t-cos\;4t}{t}$ dt = log $\sqrt{\frac{16}{36}}$ = log ( $\frac{2}{3}$ )

## Laplace Transform of Differential Equation

A differential equation with a set of initial conditions is called an initial value problem. However if the boundary conditions are given the problem is called a boundary value problem. Laplace transform serves as a useful tool in solving such problems.
We use the following procedure to solve the differential equations.
Step 1 : Write the differential equation in the form y'(t), y''(t), y'"(t) ...etc for the derivatives.
Step 2 : Take Laplace transform on both sides of the given equation.
Step 3 : We use the expressions for L[y'(t)], L[y"(t)], . . . . .
Step 4 : We substitute the given initial conditions and simplify to obtain L [ y(t) ] as a function of s.
Step 5 : We find the inverse to obtain y ( t ).

### Solved Example

Question: Solve by using Laplace Transforms$\frac{\mathrm{d^{2}}y }{\mathrm{d} t^{2}}$ + k2 y = 0 , given that y ( 0 ) = 2 and y ' ( 0 ) = 0
Solution:

The given equation is y"(t) + k2 y(t) = 0
Taking Laplace Transform on both sides, we have,

L [ y"(t) + k2 y(t) ] = L [ 0 ]

=> s2 L [y(t) - s y( 0) - y'(0) ] + k2 L [ y( t) ] = 0
Using initial conditions given, we get

( s2 + k2 ) L [ y(t) ] - 2 s = 0

=>                                           L [ y ( t ) ] = $\frac{2s}{s^{2}+k^{2}}$

=>                                                  y ( t ) = 2 L -1 [ $\frac{s}{s^{2}+k^{2}}$ ]

= 2 cos kt

Thus y ( t ) = 2 cos kt , which is the required solution.