Can Differential Equations be Solved Using Only Algebraic Techniques?

[Jhenrique]

I was reading and searching about a theory that originated the laplace transfom (http://en.wikipedia.org/wiki/Operational_calculus), but I don't know very well, second this theory, is possible to solve a differential equation only with algebric artifices (see http://es.wikipedia.org/wiki/Transformada_de_Laplace#Perspectiva_hist.C3.B3rica). But, is really possible?

Do you could show me some algebric artifice (step by step) that solve this EDO??

 

[jvicens]

Hello,

Thank you for your interest in the Laplace transform and its application in solving differential equations. The Laplace transform is indeed a powerful tool in solving differential equations, and it has been widely used in many fields of science and engineering.

To answer your question, yes, it is possible to solve a differential equation using only algebraic techniques with the help of the Laplace transform. The key idea behind the Laplace transform is to convert a differential equation into an algebraic equation, which can be easily solved using standard algebraic techniques.

Let's take an example of a simple differential equation:

y'' - 3y' + 2y = 0

To solve this using the Laplace transform, we first take the Laplace transform of both sides of the equation. The Laplace transform of a function f(t) is denoted by F(s) and is defined as:

F(s) = ∫e^(-st)f(t)dt

Applying this to our differential equation, we get:

L(y'') - 3L(y') + 2L(y) = 0

where L denotes the Laplace transform.

Now, we use the properties of the Laplace transform to simplify the equation. The Laplace transform of the first derivative is given by:

L(y') = sL(y) - y(0)

where y(0) is the initial value of y. Similarly, the Laplace transform of the second derivative is given by:

L(y'') = s^2L(y) - sy(0) - y'(0)

Substituting these values in our equation, we get:

s^2L(y) - sy(0) - y'(0) - 3(sL(y) - y(0)) + 2L(y) = 0

Simplifying this, we get:

(s^2 - 3s + 2)L(y) = (3y(0) + y'(0))

Finally, solving for L(y), we get:

L(y) = (3y(0) + y'(0)) / (s^2 - 3s + 2)

Now, to find the solution to our original differential equation, we need to take the inverse Laplace transform of L(y). This can be done using a table of Laplace transforms or by using partial fractions to simplify the expression.

I hope this helps to answer your question and gives you