In order to translate a given derivative function with a real variable t into a complex function with a variable s, the Laplace transform is used. Let f(t) be given for t ≥ 0 and suppose that the function meets some later-explained requirements.

A Complete Solution for “Laplace Transform Practice Problems”
Differential equations can be solved using the Laplace transformation method. Here, an algebraic equation in the form of a frequency domain differential equation is first converted.
To get at the differential equation’s final solution, the algebraic problem must first be solved in the frequency domain and then the answer must be converted to time domain form. In other words, the Laplace transformation can be thought of as nothing more than a quick way to solve a differential equation.
How Do You Do Laplace Transform Problems?
Differential equations can be solved using the Laplace Transform in four steps by considering applying the derivative property to the Laplace Transform of the differential equation as necessary.
First, put the resulting equation’s initial conditions in. Second, find the output variable by solving. Then, get the Laplace Transform tables’ results. At last, you must use the Inverse Laplace Transform if the result is in a format that is not represented in the tables.
Method of Laplace Transform
Several functions are subjected to the Laplace transform, including the Bessel function, unit impulse, step, unit step, shifted unit step, ramp, exponential decay, sine, cosine, hyperbolic sine, and hyperbolic cosine. The ability to quickly solve higher order differential equations by transforming them into algebraic equations is the Laplace transform’s biggest benefit, though.
To perform a Laplace transform of a time function, a set of procedures must be followed. The steps that must be taken in order to convert a given function of time f(t) into its corresponding Laplace transform are first f(t)is multiplied by e-st, where s is a complex number (s = σ + j ω).
Then, integrate this product with respect to time using zero and infinity as limits. F(s) represents the Laplace transformation of f(t) as a result of this integration (s). The method that is used to recover the time function f(t) from the Laplace transform is known as the inverse Laplace transformation and it is symbolized by the symbol £-1.

Table of Laplace Transform
There are some basic transforms that help us in solving complex functions. Here is the table of such Laplace transforms given below-

How Do Laplace Transforms Work?
Numerous characteristics of the Laplace transform make it effective for studying linear dynamical systems. The biggest benefit is that by s, integration becomes division and differentiation becomes multiplication (evocative of the way logarithms alter multiplication to addition of logarithms).
The Laplace variable s is also referred to as an operator variable in the L domain because of this characteristic, which can refer to either a derivative operator or (for s−1) an integration operator. Integral and differential equations are transformed into considerably simpler polynomial equations, making them much easier to solve. The inverse Laplace transform returns to the original domain after being solved.
What Is the Laplace Transform of e’t 2?
A naive attempt to calculate the Laplace transform of the function f(t)=et2f(t)=et2 results in integrals of the form,

which obviously don’t exist as the integrand grows too large. However, Wolfram Alpha/Mathematica gives the following result,

where F is the Dawson Function.
The Laplace transform of this function is,

for every real number s. Thus, the function does not have a Laplace transform.
What Is E-st in Laplace Transform?
A time domain function is transformed using the Laplace transform by integrating from 0 to infinity multiplied by e-st, of the time domain function. Here it is an exponential function where t denotes the function of time and s denotes a complex number that is s = σ + j ω.
Conclusion
A differential equation can be reduced via the Laplace transform into an easy algebraic problem. Algebra is still simpler to solve than a differential equation, even as it gets a little more complicated. However, this approach can only be used to resolve differential equations whose constants are known.
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