please follow this link,
u can see its brief advantage compared to the usual method
http://books.Google.com/books?id=zMcAXrJpyPkC&pg=PA902&lpg=PA902&dq=advantages+of+laplace+transform+for+solving+differential+equations&source=bl&ots=txzL6fkRMR&sig=rFigMIeYaT3T65q-ydLM8kjioyE&hl=en&ei=KQ3kSrr5CILA-Qann4nJCQ&sa=X&oi=book_result&ct=result&resnum=8&ved=0CCMQ6AEwBw#v=onepage&q=&f=false
A Laplace transform is a mathematical operator that is used to solve differential equations. This operator is also used to transform waveform functions from the time domain to the frequency domain and can simplify the study of such functions. For continuous functions, f(t), the Laplace transform, F(s), is defined as the Integral from 0 to infinity of f(t)*e-stdt. When this definition is used it can be shown that the Laplace transform, Fn(s) of the nth derivative of a function, fn(t), is given by the following generic formula:Fn(s)=snF(s) - sn-1f0(0) - sn-2f1(0) - sn-3f2(0) - sn-4f3(0) - sn-5f4(0). . . . . - sn-nfn-1(0)Thus, by taking the Laplace transform of an entire differential equation you can eliminate the derivatives of functions with respect to t in the equation replacing them with a Laplace transform operator, and simple initial condition constants, fn(0), times a new variable s raised to some power. In this manner the differential equation is transformed into an algebraic equation with an F(s) term. After solving this new algebraic equation for F(s) you can take the inverse Laplace transform of the entire equation. Since the inverse Laplace transform of F(s) is f(t) you are left with the solution to the original differential equation.
Laplace will only generate an exact answer if initial conditions are provided
the Laplace transform of sin2 3thttp://www7.0zz0.com/2009/12/30/19/748450027.gif
s
2/s
The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes ofvibration (frequencies), the Laplace transform resolves a function into its moments. Like the Fourier transform, the Laplace transform is used for solving differential and integral equations.
Yes, the Laplace transform can be used to solve differential equations without initial values, but the process is somewhat different. Typically, the Laplace transform is applied to linear ordinary differential equations with given initial conditions to find solutions in the transformed domain. When initial values are not specified, the solution may involve arbitrary constants, representing a family of solutions that satisfy the differential equation. These constants can often be determined by additional conditions or constraints imposed on the problem.
Laplace transforms to reduce a differential equation to an algebra problem. Engineers often must solve difficult differential equations and this is one nice way of doing it.
yes
with the help of laplace transform the calculation part can be reduced in frequency domain .In time domain differential equations are used and solution is cumbersome.
you apply the Laplace transform on both sides of both equations. You will then get a sytem of algebraic equations which you can solve them simultaneously by purely algebraic methods. Then take the inverse Laplace transform .
Laplace Transforms are used to solve differential equations.
The Laplace transform is a powerful mathematical tool used to solve partial differential equations (PDEs) by transforming them into algebraic equations in the Laplace domain. It simplifies the process of handling initial and boundary value problems by converting derivatives with respect to time into polynomial terms. Once the PDE is solved in the transformed domain, the inverse Laplace transform is applied to revert to the original function, providing solutions in the time domain. This technique is particularly useful for linear PDEs with constant coefficients and can facilitate the analysis of systems in engineering and physics.
Some differential equations can become a simple algebra problem. Take the Laplace transforms, then just rearrange to isolate the transformed function, then look up the reverse transform to find the solution.
The S transform in circuit analysis and design is method for transforming the differential equations describing a circuit in terms of dt into differential equations describing a circuit in terms of ds. With t representing the time domain and s representing the frequency domain.Usually the writing of the time domain equations for the circuit is skipped and the circuit is redrawn in the frequency domain first and the equations are taken directly from this transformed circuit. This is actually much simpler and faster than transforming the time domain equations of the circuit would be.The S transform and Laplace transform are related operations but different; the S transform operates on circuits and describes how they modify signals, the Laplace transform operates on signals.
In short, yes, it is possible, but much, much more difficult. Laplace transforms turn systems of integro-differential equations into algebraic equations, and give an immediate expression for the frequency response which is very heavily used in design.
Laplace Transformation is modern technique to solve higher order differential equations.It has several great advantages over old classical method, such as: # In this method we don't have to put the values of constants by our self. # We can solve higher order differential equations also of more than second degree equations because using classical mothed we can only solve first or second degree differential equations.