Diff eq - LT Discontinuous Sources HW

In summary, the conversation discusses the partial fraction method and its disadvantages in terms of time and complexity. The conversation also explores alternative methods, such as finding the inverse of the function, to make the process easier. The identities for partial fractions are mentioned as a helpful tool in solving the problem. The person asking the question expresses their gratitude for the advice given and shares that it has been successful in solving some questions. The person offering the advice acknowledges that it may not work for all problems, but when it does, it simplifies the process significantly.
  • #1
Arij
25
2

Homework Statement


15whzic.png


Homework Equations


2j495qh.png

The Attempt at a Solution


this is my attempt, I figured this would be easier than typing.
2hhipmq.jpg


any how, I find partial fraction method is so time consuming and very algebraically complicated, is there a trick or another method I could use to make this easier?

thanks
 
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  • #2
Arij said:

Homework Statement


15whzic.png


Homework Equations


2j495qh.png

The Attempt at a Solution


this is my attempt, I figured this would be easier than typing.
2hhipmq.jpg


any how, I find partial fraction method is so time consuming and very algebraically complicated, is there a trick or another method I could use to make this easier?

thanks

Convert ##Y_1(s) = \frac{1}{(s+3)(s-4)(s-8)}## to partial fractions, then multiply by ##5 e^{-9s}## later; better still, find the inverse of ##Y_1(s)## as ##y_1(t)##; the solution is then ##5 u(t-9)y_1(t-9)##.

The partial fractions can be found easily (but still with some algebra) by applying successively the identities
$$\frac{1}{(s-a)(s-b)} = \frac{1}{a-b} \left[ \frac{1}{s-a} - \frac{1}{s-b} \right] \: \text{if} \; a \neq b$$
 
  • #3
Ray Vickson said:
Convert ##Y_1(s) = \frac{1}{(s+3)(s-4)(s-8)}## to partial fractions, then multiply by ##5 e^{-9s}## later; better still, find the inverse of ##Y_1(s)## as ##y_1(t)##; the solution is then ##5 u(t-9)y_1(t-9)##.

The partial fractions can be found easily (but still with some algebra) by applying successively the identities
$$\frac{1}{(s-a)(s-b)} = \frac{1}{a-b} \left[ \frac{1}{s-a} - \frac{1}{s-b} \right] \: \text{if} \; a \neq b$$
That was so much help
Ray Vickson said:
5u(t−9)y1(t−9
I used this and worked like miracles for few questions, I was just confused of changing to and from the t-c.
 
  • #4
Arij said:
That was so much help

*************************************************
I am glad. It does not always work, but when it does it makes life a lot easier.
*************************************************


I used this and worked like miracles for few questions, I was just confused of changing to and from the t-c.
 

Related to Diff eq - LT Discontinuous Sources HW

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model many physical, biological, and economic phenomena.

2. What is the Laplace transform?

The Laplace transform is an integral transform that converts a function of time into a function of complex frequency. It is often used to solve differential equations with discontinuous sources.

3. What are discontinuous sources in differential equations?

Discontinuous sources in differential equations refer to functions that are not continuous over a certain interval. These sources can cause sudden changes or disruptions in the system being modeled.

4. How do you solve differential equations with discontinuous sources using Laplace transform?

To solve differential equations with discontinuous sources using Laplace transform, the Laplace transform of the equation is taken and the discontinuous sources are treated as initial conditions. Then, the inverse Laplace transform is applied to obtain the solution in the time domain.

5. What are some real-world applications of differential equations with discontinuous sources?

Differential equations with discontinuous sources are commonly used to model systems with sudden changes or disruptions, such as electrical circuits, mechanical systems, and chemical reactions. They can also be used in economics to model sudden changes in demand or supply.

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