inglcorn quyteri tareek

[Amie Mandy]

We are going to be given a transform, \(F(s)\), and ask what function (or functions) did we have originally. As you will see this can be a more complicated and lengthy process than taking transforms. In these cases we say that we are finding the Inverse Laplace Transform of \(F(s)\) and use the following notation.

The denominator of the third term appears to be #3 in the table with \(n = 4\). The numerator however, is not correct for this. There is currently a 7 in the numerator and we need a \(4! = 24\) in the numerator. This is very easy to fix. Whenever a numerator is off by a multiplicative constant, as in this case, all we need to do is put the constant that we need in the numerator. We will just need to remember to take it back out by dividing by the same constant.

So, probably the best way to identify the transform is by looking at the denominator. If there is more than one possibility use the numerator to identify the correct one. Fix up the numerator if needed to get it into the form needed for the inverse transform process. Finally, take the inverse transform.

Recall that in completing the square you take half the coefficient of the \(s\), square this, and then add and subtract the result to the polynomial. After doing this the first three terms should factor as a perfect square.

We needed an \(s + 4\) in the numerator, so we put that in. We just needed to make sure and take the 4 back out by subtracting it back out. Also, because of the 3 multiplying the \(s\) we needed to do all this inside a set of parenthesis. Then we partially multiplied the 3 through the second term and combined the constants. With the transform in this form, we can break it up into two transforms each of which are in the tables and so we can do inverse transforms on them,

The last part of this example needed partial fractions to get the inverse transform. When we finally get back to differential equations and we start using Laplace transforms to solve them, you will quickly come to understand that partial fractions are a fact of life in these problems. Almost every problem will require partial fractions to one degree or another.

In order for these two to be equal the coefficients of the \(s^2\), \(s\) and the constants must all be equal. So, setting coefficients equal gives the following system of equations that can be solved.

Laplace transform is named in honour of the great French mathematician, Pierre Simon De Laplace (1749-1827). Like all transforms, the Laplace transform changes one signal into another according to some fixed set of rules or equations. The best way to convert differential equations into algebraic equations is the use of Laplace transformation.

Laplace transformation plays a major role in control system engineering. To analyze the control system, Laplace transforms of different functions have to be carried out. Both the properties of the Laplace transform and the inverse Laplace transformation are used in analyzing the dynamic control system. In this article, we will discuss in detail the definition of Laplace transform, its formula, properties, Laplace transform table and its applications in a detailed way.

A function is said to be a piecewise continuous function if it has a finite number of breaks and it does not blow up to infinity anywhere. Let us assume that the function f(t) is a piecewise continuous function, then f(t) is defined using the Laplace transform. The Laplace transform of a function is represented by Lf(t) or F(s). Laplace transform helps to solve the differential equations, where it reduces the differential equation into an algebraic problem.

The Laplace transform is a well established mathematical technique for solving a differential equation. Many mathematical problems are solved using transformations. The idea is to transform the problem into another problem that is easier to solve. On the other side, the inverse transform is helpful to calculate the solution to the given problem.

The Laplace transform can also be defined as bilateral Laplace transform. This is also known as two-sided Laplace transform, which can be performed by extending the limits of integration to be the entire real axis. Hence, the common unilateral Laplace transform becomes a special case of Bilateral Laplace transform, where the function definition is transformed is multiplied by the Heaviside step function.

In pure and applied probability theory, the Laplace transform is defined as the expected value. If X is the random variable with probability density function, say f, then the Laplace transform of f is given as the expectation of:

Examples of how to use Laplace transform to solve ordinary differential equations (ODE) are presented. One of the main advantages in using Laplace transform to solve differential equations is that the Laplace transform converts a differential equation into an algebraic equation.
Heavy calculations involving decomposition into partial fractions are presented in the appendix at the bottom of the page.

Example 4
Use Laplace transform to solve the differential equation\[ y'' - y' - 2 y = \sin(3t) \]with the initial conditions \( y(0) = 1 \) and \( y'(0) = -1 \).
Solution to Example 4
Let \( Y(s) \) be the Laplace transform of \( y(t) \)
Take the Laplace transform of both sides of the given differential equation
\( \mathscrL\ y'' - y' - 2 y \ = \mathscrL\ \sin(3t) \ \)
Use linearity property of Laplace transform to expand the left side and use table to evaluate the right side.
\( \mathscrL\ y"\ - \mathscrL\ y'\ - 2 \mathscrL\ y \ = \dfrac3s^2+3^2 \)
Use first and second derivative properties to rewrite the terms \( \mathscrL\ y"\ \) and \( \mathscrL\ y'\ \) and simplify the right side.
\( s^2 Y(s) - s y(0) - y'(0) - (sY(s) - y(0)) + 2 Y(s) = \dfrac3s^2+3^2 \)
Substitute \( y(0) \) and \( y'(0) \) by their numerical values and expand
\( s^2 Y(s) - s + 1 - s Y(s) + 1 - 2 Y(s) = \dfrac3s^2+3^2 \)
Group like terms and keep terms with \( Y(s) \) on the left side of the equation
\( s^2 Y(s) - s Y(s) - 2 Y(s) = \dfrac3s^2+3^2 + s - 2 \)
Factor \( Y(s) \) out on the left side
\( Y(s) (s^2 - s - 2 ) = \dfrac3s^2+3^2 + s - 2 \)
Solve the above for \( Y(s) \)
\( Y(s) = \dfrac3(s^2+3^2)(s^2 - s - 2) + \dfracs-2s^2 - s - 2 \)
Factor the term \( s^2 - s - 2 \) in the denominator
\( Y(s) = \dfrac3(s^2+3^2)(s-2)(s+1) + \dfracs-2(s-2)(s+1) \)
which may be expanded in partial fractions as (see Appendix C at the bottom of the page for details).
\( Y(s) = \dfrac3s130(s^2+3^2) - \dfrac33130(s^2+3^2) + \dfrac910(s+1) + \dfrac113(s-2) \)
We now use formulas in the table of formulas of Laplace tranform to find the inverse Laplace transform of \( Y(s) \) which is given by
\( y(t) = \dfrac3130 \cos(3x) - \dfrac11130 \sin(3x) + \dfrac910 e^-x +\dfrac113 e^2x\)

Expand in partial fractions from example 4
\( \dfrac3(s^2+3^2)(s^2 - s - 2) + \dfracs-2s^2 - s - 2\)
Factor denominators
\( \dfrac3(s^2+3^2)(s-2)(s+1) + \dfracs-2(s-2)(s+1)\)
Simplify the term on the right
\( \dfrac3(s^2+3^2)(s-2)(s+1) + \dfrac1s+1 \)
Express into partial fractions
\( \dfrac3(s^2+3^2)(s-2)(s+1) + \dfrac1s+1 = \dfracAs + Bs^2+3^2 + \dfracCs+1 + \dfracDs-2 \)
Multiply all terms of the above by the denominator \( (s^2+3^2)(s-2)(s+1) \) and simplify
\( 3 + (s^2+3^2)(s-2) = (As + B)(s-2)(s+1) + C (s^2+3^2)(s-2) + D (s^2+3^2)(s+1) \) (1)
Select values of \( s \) that simplify calculations for the coefficients \( A, B, C \) and \( D \)
Set \( s = 2 \) on both sides of equation (1)
\( 3 + (2^2+3^2)(2-2) = (2 A + B)(2-2)(s+1) + C (2^2+3^2)(2-2) + D (2^2+3^2)(2+1) \)
Simplify
\( 3 = 39 D \)
Solve for \( D \)
\( D = \dfrac113 \)
Set \( s = -1 \) on both sides of equation (1)
\( 3 + ((-1)^2+3^2)(-1-2) = (-A + B)(-1-2)(-1+1) + C ((-1)^2+3^2)(-1-2) + D ((-1)^2+3^2)(-1+1) \)
Simplify
\( 3 - 30 = - 30 C \)
Solve for \( C \)
\( C = \dfrac910 \)
Set \( s = 0 \) on both sides of equation (1)
\( 3 +(0^2+3^2)(0-2) = (0 + B)(0-2)(0+1) + C (0^2+3^2)(0-2) + D (0^2+3^2)(0+1) \)
Simplify
\( 3 - 18 = -2 B - 19 C + 9D \)
Substitute \( C \) and \( D \) by their numerical values obtained above and solve for B to obtain
\( B = -\dfrac33130 \)
Set \( s = 1 \) on both sides of equation (1)
\( 3 + (1^2+3^2)(1-2) = (A + B)(1-2)(1+1) + C (1^2+3^2)(1-2) + D (1^2+3^2)(1+1) \)
Substitute \( B, C \) and \( D \) by their numerical values obtained above and solve for A to obtain
\( A = \dfrac3130 \)
Hence
\( \dfrac3(s^2+3^2)(s^2 - s - 2) + \dfracs-2s^2 - s - 2\)

\( \quad \quad = \dfracAss^2+3^2 + \dfracBs^2+3^2 + \dfracCs+1 + \dfracDs-2 \)

\( \quad \quad = \dfrac 3s130(s^2+3^2) - \dfrac33130(s^2+3^2) + \dfrac910(s+1) + \dfrac113(s-2) \)

In undergraduate differential equations it's usual to deal with the Laplace transform to reduce the differential equation problem to an algebraic problem.The Laplace transform of a function $f(t)$, for $t \geq 0$ is defined by $\int_0^\infty f(t) e^-st dt$.How to avoid looking at this definition as "magical"? How to somehow discover it from more basic definitions?

This answer is not exactly an answer to the original question, but this is for the benefit of MO user vonjd who wanted to know more details about the similarities between solving differential equations through Laplace transforms and solving recurrence relations using generating functions.

I will do an example of each, and this should be enough to show the similarities. In each case, we have a linear equation with constant coefficients; this is where both methods really shine, although they both can handle some variable coefficients more or less gracefully. Ultimately, the biggest challenge is to apply the inverse transform: always possible in the linear case, not so easy otherwise.

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