100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Sumario APLICACION DE LA TRANSFORMADA DE LAPLACE $2.99   Add to cart

Summary

Sumario APLICACION DE LA TRANSFORMADA DE LAPLACE

 8 views  0 purchase
  • Course
  • Institution
  • Book

Este PDF proporciona unas aplicaciones de la Transformada de Laplace, cubriendo sus definiciones básicas, propiedades fundamentales y aplicaciones prácticas en la resolución de ecuaciones diferenciales. Ideal para estudiantes y profesionales que buscan una comprensión rápida y efectiva de este...

[Show more]

Preview 2 out of 12  pages

  • No
  • Unknown
  • June 23, 2024
  • 12
  • 2022/2023
  • Summary
avatar-seller
Esta es la demostración de la fórmula:

Usemos las siguientes formulas:

2 1
𝑀(𝑎) = ∫ 𝑒 −𝑎𝑥 sin⁡(𝑏𝑥)(πcoth(𝜋𝑥) − )𝑑𝑥
0 𝑥

1 2𝑥
𝜋 coth(𝜋𝑥) − = ∑ 2
𝑥 𝑥 + 𝑘2
𝑘=1

Hallemos la transformada de laplace:

𝐿 = ∫ 𝑒 −𝑎𝑐 𝑀(𝑎)𝑑𝑎
0
∞ ∞
2 1
𝐿 = ∫ ∫ 𝑒 −𝑎𝑐 𝑒 −𝑎𝑥 sin⁡(𝑏𝑥)(πcoth(𝜋𝑥) − )𝑑𝑥 𝑑𝑎
0 0 𝑥

Integramos con respecto a ‘a’:

1 1
𝐿=∫ (sin⁡(𝑏𝑥)(πcoth(𝜋𝑥) − ))𝑑𝑥
0 𝑥2 + 𝑐 𝑥

Usamos la serie infinita:

1 2𝑥
𝜋 coth(𝜋𝑥) = + ∑ 2
𝑥 𝑥 + 𝑘2
𝑘=1

∞ ∞
2𝑥 sin⁡(𝑏𝑥)𝑑𝑥
𝐿 = ∫ (∑ )
0 𝑥2 +𝑘 2 𝑥2 + 𝑐
𝑘=1

Separamos en fracciones parciales:
∞ ∞
2xsin⁡(𝑏𝑥)𝑑𝑥
𝐿 = ∑∫
(𝑥 2 + 𝑐)(𝑥 2 + 𝑘 2 )
𝑘=1 0
∞ ∞
2 xsin(𝑏𝑥) 1 1
𝐿 = ∑∫ ( 2 − 2 )𝑑𝑥
(𝑘 − 𝑐) (𝑥 + 𝑐) (𝑥 + 𝑘 2 )
2
𝑘=1 0

Seguimos separando:
∞ ∞ ∞ ∞
2 xsin(𝑏𝑥) 1 2 xsin(𝑏𝑥) 1
𝐿 = ∑∫ 2
( 2
) 𝑑𝑥 − ∑ ∫ 𝑑𝑥
0 (𝑘 − 𝑐) (𝑥 + 𝑐) 0 (𝑘 − 𝑐) (𝑥 + 𝑘 2 )
2 2
𝑘=1 𝑘=1

, Aquí, usamos esta fórmula:

xsen⁡(𝑏𝑥)𝑑𝑥 𝜋𝑒 −𝑎𝑏
∫ =
0 𝑥 2 + 𝑎2 2

Aplicamos a la transformada:
∞ ∞ ∞ ∞
2 xsin(𝑏𝑥) 1 2 xsin(𝑏𝑥) 1
𝐿 = ∑∫ (
2 − 𝑐) (𝑥 2 + 𝑐)
) 𝑑𝑥 − ∑ ∫ 2 − 𝑐) (𝑥 2 + 𝑘 2 )
𝑑𝑥
0 (𝑘 0 (𝑘
𝑘=1 𝑘=1
∞ ∞
𝜋𝑒 −√𝑐𝑏 𝜋𝑒 −𝑘𝑏
𝐿 = 2∑ − 2 ∑
2(𝑘 2 − 𝑐) 2(𝑘 2 − 𝑐)
𝑘=1 𝑘=1
∞ ∞
−√𝑐𝑏
1 𝑒 −𝑘𝑏
𝐿 = 𝜋𝑒 ∑ 2 −𝜋 ∑ 2
(𝑘 − 𝑐) (𝑘 − 𝑐)
𝑘=1 𝑘=1

La suma resaltada, lo reemplazamos con esta fórmula:

1 2𝑥
𝜋 coth(𝜋𝑥) − = ∑ 2
𝑥 𝑥 + 𝑘2
𝑘=1

Haciendo

𝑥 = 𝑖 √𝑐

1 2𝑖 √𝑐
𝜋 coth(𝜋𝑖 √𝑐) − =∑
𝑖 √𝑐 −𝑐 + 𝑘 2
𝑘=1

1 𝜋 cot(𝜋𝑖 √𝑐) 1
∑ =− +
(𝑘 2 − 𝑐) 2√𝑐 2𝑐
𝑘=1

Y reemplazamos en ‘L’:

𝜋 cot(𝜋√𝑐) 1 𝑒 −𝑘𝑏
𝐿= 𝜋𝑒 −√𝑐𝑏 (− + )−𝜋∑ 2
2√𝑐 2𝑐 (𝑘 − 𝑐)
𝑘=1

Se viene lo chido, calculemos la transformada inversa de L, para aplicar el teorema de unicidad
para funciones continuas de Laplace, esto dice:

Si dos funciones tienen igual transformada de Laplace, entonces son iguales:

Primero, demonos cuenta de lo siguiente:

𝑒 2𝜋√𝑐𝑖 + 1
cot(𝜋√𝑐) = (𝑖)
(𝑒 2𝜋√𝑐𝑖 − 1)
Por identidad de Euler, un viejo conocido:

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller jesussheldonmontgomeryerdos. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $2.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

62890 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$2.99
  • (0)
  Add to cart