# Laplace transform of derivative and integro- differential equation

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### Parul Institute of Engineering & Technology:

Parul Institute of Engineering & Technology Subject Code : 130001 Name Of Subject : Mathematics-3 Name of Unit : Laplace Transformation Topic : Transform of derivative and integral differential equation Name of Faculty : Shraddha Madam Name of Students: (1) Rahul M Valvi (2) Sourin Patel

### Laplace Transform of Derivative :

Laplace Transform of Derivative

### Laplace transform of first derivative of function f(t):

Laplace transform of first derivative of function f(t) Suppose that we have a function f(t) such that it is continuos for time t ≥ 0, then Laplace transform of derivative of f(t) i.e., f’(t) exists and …………(1) where F(s)= L{f(t)} and f(0) is value of function at time t=0 L {f’(t)} = s F(s) - f(0)

### Laplace transform of higher derivatives of function f(t):

Laplace transform of higher derivatives of function f(t) Let f(t),f’(t),f”(t)………f ͫ(t) be continous for time t ≥ 0 then laplace transform of higher order derivative exists and is given by ……………..(2) L { f (n ) ( t ) } = s ( n ) F ( s ) - s ( n-1 ) f (0) - s (n-2) f’(0)..........s (0) f (n-1) (0)

### Hence Laplace transform of various derivatives of function f(t) are:

Hence Laplace transform of various derivatives of function f(t) are . 1 . L { f ‘ ( t ) } = s F ( s ) - f (0) 2 . L { f “ ( t ) } = s 2 F ( s ) - s f (0) - f’(0) 3 . L { f “’ ( t ) } = s 3 F ( s ) - s 2 f (0) – s f’(0) - f “ (0) 4. L { f (4 ) ( t ) } = s ( 4 ) F ( s ) - s ( 3 ) f (0) - s (2) f’(0) – s f” (0) - f “’ (0) and so on for higher order derivatives

### Now, we will take few examples on laplace transform of derivative :

Now, we will take few examples on laplace transform of derivative 1. Let f(t)= sin (at) . Find its Laplace transform by utilizing the concept of laplace transform of a derivative Solution : f(t) = sin(at); f(0) = 0 ; f’(t) = a cos (at); f’(0) = a; f”(t) = -a 2 sin(at) and f”(0) = 0 Now we know Laplace tranform of f ”(t) is given by the formula L { f “ ( t ) } = s 2 F ( s ) - s f (0) - f’(0) = s 2 F ( s )- s*0 – a = s 2 F ( s )- a …………………..(1) Also, L{f”(t)} = L(-a 2 sin(at) ) = -a 2 L(sin(at)} = -a 2 F(s)………………....(2) Comparing (1) and (2) we get -a 2 F(s) = s 2 F ( s )- a So, F(s) = a/(s 2 + a 2 ) Hence the answer.

### PowerPoint Presentation:

2. Let f(t) = tsin (at) . Find Laplace transform of f(t) Solution : f(t) = tsin (at) ; f(0) = 0 f’(t) = atcos (at) + sin(at) f’(0) = 0 f”(t) = 2acos(at) – a 2 tsin(at) f”(0) = 2a = 2acos(at) – a 2 f(t) ……………………………………………………….(1) Now Laplace transform of second order derivative is given by L{f”(t)} = L{2acos(at)- a 2 f(t)} = 2a L { cos (at)} – a 2 L {f(t)} = 2a (s/(s 2 + a 2 )) – a 2 F(s)…………………………………………..(2) Also, L { f “ ( t ) } = s 2 F ( s ) - s f (0) - f’(0) = s 2 F ( s ) – s*0 - 0 ………..from (1) = s 2 F ( s ) ……………………………………………...(3)

### PowerPoint Presentation:

Comparing (2) and (3) we get 2a (s/(s 2 + a 2 )) – a 2 F(s) = s 2 F ( s ) So, F(s) = 2as/(s 2 + a 2 ) Hence the answer

### Integro- Differential Equations:

Integro - Differential Equations

### PowerPoint Presentation:

An integro -differential equation is an equation which involves both integrals and derivatives of a function. The general first-order, linear integro -differential equation is of the form ….. As is typical with differential equations, obtaining a closed-form solution can often be difficult. In the relatively few cases where a solution can be found, it is often by some kind of integral transform, where the problem is first transformed into an algebraic setting. In such situations, the solution of the problem may be derived by applying the inverse transform to the solution of this algebraic equation.

### PowerPoint Presentation:

Consider the following first-order problem, The Laplace transform is defined by, Upon taking term-by-term Laplace transforms, and utilising the rules for derivatives and integrals, the integro -differential equation is converted into the following algebraic equation, Inverting the Laplace transform using contour integral methods then gives ,