Summary of "Lecture 36-Applications of Laplace Transforms-I"

Summary of Lecture 36 - Applications of Laplace Transforms - Part I

This lecture focuses on applying Laplace transforms to solve various types of differential equations commonly encountered in engineering problems, particularly electrical circuits. It demonstrates step-by-step methodologies for solving ordinary differential equations (ODEs) and integral equations involving discontinuous functions using Laplace transforms.

Main Ideas and Concepts

Applications of Laplace Transforms:
- Solving ordinary differential equations (ODEs) in engineering (electrical, mechanical).
- Handling integral equations.
- Managing equations involving discontinuous functions.
General Approach:
- Take the Laplace transform of both sides of the differential equation.
- Use initial conditions to simplify the transformed equation.
- Solve for the Laplace transform of the unknown function.
- Find the inverse Laplace transform to get the solution in the time domain.

Detailed Methodologies and Examples

1. Solving an Electrical Circuit Differential Equation (Current in L-R Circuit)

Problem:

[ L \frac{di}{dt} + Ri = E \sin(\Omega t) ]

Initial condition: ( i(0) = 0 )
Find current ( i(t) ) at any time ( t ).

Steps:

Take Laplace transform of both sides.
Use property: [ \mathcal{L}\left{ \frac{di}{dt} \right} = pI(p) - i(0) ]
Substitute initial conditions and known Laplace transforms, e.g., [ \mathcal{L}{\sin \Omega t} = \frac{\Omega}{p^2 + \Omega^2} ]
Solve algebraically for ( I(p) ).
Use partial fraction decomposition or convolution theorem to find inverse Laplace transform.
The solution has the form: [ i(t) = a \cos(\Omega t) + \frac{b}{\Omega} \sin(\Omega t) + \frac{c}{L} e^{-\frac{R}{L}t} ]
Constants ( a, b, c ) found by comparing coefficients.

2. Electrical Circuit with Switching (Switch Connected at ( t=0 ), Disconnected at ( t=a ))

Problem:

Same circuit equation as above.
EMF ( E ) is applied from ( t=0 ) to ( t=a ), then zero afterward.
Find ( i(t) ) for any ( t ).

Steps:

Express EMF as a piecewise function.
Take Laplace transform of the piecewise EMF using integration limits.
Use initial condition ( i(0) = 0 ).
Solve for ( I(p) ) involving terms with ( e^{-ap} ).
Use partial fraction decomposition to simplify.
Inverse Laplace transform results in piecewise time-domain solution reflecting switch operation:
- For ( 0 \leq t < a ), current builds up.
- For ( t \geq a ), current decays.

3. Solving a Second Order ODE with Exponential Forcing

Problem:

[ y’’ - 6y’ + 5y = e^{2t} ]

Initial conditions: ( y(0) = 1 ), ( y’(0) = -1 ).

Steps:

Take Laplace transform of both sides.
Use Laplace transform of derivatives: [ \mathcal{L}{y’} = pY(p) - y(0), \quad \mathcal{L}{y’‘} = p^2 Y(p) - p y(0) - y’(0) ]
Substitute initial conditions.
Rearrange to solve for ( Y(p) ).
Use partial fractions to decompose ( Y(p) ).
Take inverse Laplace transform to get ( y(t) ).

4. Problem Involving ( t y’(t) ) Terms

Problem:

[ t y’ + 2 y’ + t y = 0 ]

Boundary conditions: ( y(0) = 1 ), ( y(\pi) = 0 ).

Steps:

Use Laplace transform properties involving multiplication by ( t ): [ \mathcal{L}{ t f’(t) } = -\frac{d}{dp} \left[ p F(p) - f(0) \right] ]
Take Laplace transform of each term.
Form differential equation in ( p )-domain.
Solve for ( F(p) ).
Use initial/boundary conditions and corollary of existence theorem to find constants.
Find inverse Laplace transform, which leads to: [ y(t) = \frac{\sin t}{t} ]
Verify ( y(\pi) = 0 ) satisfies the condition.

Key Laplace Transform Properties Used

( \mathcal{L}{ f’(t) } = p F(p) - f(0) )
( \mathcal{L}{ f’‘(t) } = p^2 F(p) - p f(0) - f’(0) )
( \mathcal{L}{ \sin(\omega t) } = \frac{\omega}{p^2 + \omega^2} )
( \mathcal{L}{ e^{at} } = \frac{1}{p - a} )
Multiplication by ( t ) in time domain corresponds to differentiation in Laplace domain: [ \mathcal{L}{ t f(t) } = - \frac{d}{dp} F(p) ]

Summary of the Process for Solving ODEs with Laplace Transforms

Formulate the differential equation and initial/boundary conditions.
Apply Laplace transform to each term of the equation.
Use initial conditions to simplify transformed terms.
Solve the resulting algebraic equation for the Laplace transform of the unknown function.
Use partial fraction decomposition or convolution theorem to simplify the expression.
Find the inverse Laplace transform to obtain the solution in the time domain.
Verify the solution satisfies initial/boundary conditions.

Speakers/Sources Featured

Main Speaker: The lecturer delivering the mathematical lecture on Laplace transforms and their applications (name not specified).
Background: The lecture is part of a series on mathematical methods and applications, likely from an academic or educational institution.

Additional Notes

The lecture emphasizes using Laplace transforms as a powerful alternative to classical methods for solving differential equations.
It demonstrates the handling of piecewise and discontinuous functions using Laplace transforms.
The examples focus primarily on electrical circuit problems but also include general ODEs with exponential and trigonometric forcing functions.
The lecture encourages familiarity with Laplace transform properties, partial fractions, and inverse transforms for practical problem-solving.

This summary captures the core methodologies, example problems, and key concepts presented in the lecture on applications of Laplace transforms.