Chapter 10: Problem 5
Use \((10.10 .1)\) to determine \(L[f]\). $$f(t)=7 t e^{-t}$$
Short Answer
Expert verified
The Laplace transform of \(f(t) = 7te^{-t}\) is given by \(L[f(t)] = \frac{7}{(s+1)^2}\).
Step by step solution
01
Recall the Laplace Transform definition
The Laplace transform of a function is defined as:
\(L[f(t)] = F(s) = \int_0^\infty e^{-st}f(t)dt\)
For the given function \(f(t) = 7te^{-t}\), we will apply this definition to compute the Laplace transform.
02
Set up the integral
Substitute the given function into the Laplace Transform definition:
\(F(s) = \int_0^\infty e^{-st}(7te^{-t})dt\)
03
Simplify the integral
Combine the exponentials inside the integral:
\(F(s) = 7\int_0^\infty te^{-(s+1)t}dt\)
04
Integration by Parts
Now we need to integrate by parts to find the solution to this integral. We will take the following:
\(u = t, dv = e^{-(s+1)t}dt\)
Differentiate u and integrate dv to find du and v:
\(du = dt\)
\(v = -\frac{1}{s+1}e^{-(s+1)t}\)
Now, apply integration by parts:
\(F(s) = 7[-t\frac{1}{s+1}e^{-(s+1)t} \Big|_0^\infty + \int_0^\infty \frac{1}{s+1}e^{-(s+1)t}dt\)
05
Evaluate the remaining integral
Evaluate the remaining integral using the initial definition of Laplace Transform:
\(F(s) = 7[-t\frac{1}{s+1}e^{-(s+1)t}\Big|_0^\infty - \frac{1}{(s+1)^2}L[e^{-(s+1)t}]\)
\(F(s) = 7[-t\frac{1}{s+1}e^{-(s+1)t}\Big|_0^\infty - \frac{1}{(s+1)^2}\)
06
Evaluate the limit
Evaluate the term within the brackets as t approaches infinity. Using L'Hopital's rule if needed or notice that exponential will dominate and approach 0 quicker than t:
\(F(s) = 7[-(0 - \frac{1}{s+1}) - \frac{1}{(s+1)^2}\)
07
Simplify the result
Simplify the Laplace transform result:
\(F(s) = \frac{7}{(s+1)^2}\)
The Laplace transform of the given function is: \(L[f(t)] = \frac{7}{(s+1)^2}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a powerful technique used to solve integrals, especially when the integrand is a product of two functions. The basic idea is to transform a complex integral into simpler parts that can be solved more easily.
Suppose you have two functions, say \(u\) and \(dv\). Integration by parts works on the principle of the formula:
In the given exercise, the function \(f(t) = 7te^{-t}\) was addressed using integration by parts. Here, \(u = t\) and \(dv = e^{-(s+1)t}dt\), which implies differentiating \(u\) to get \(du = dt\), and integrating \(dv\) to get \(v = -\frac{1}{s+1}e^{-(s+1)t}\). This transforms the original integral into manageable components, allowing further simplification.
Suppose you have two functions, say \(u\) and \(dv\). Integration by parts works on the principle of the formula:
- \(\int u \, dv = uv - \int v \, du\)
In the given exercise, the function \(f(t) = 7te^{-t}\) was addressed using integration by parts. Here, \(u = t\) and \(dv = e^{-(s+1)t}dt\), which implies differentiating \(u\) to get \(du = dt\), and integrating \(dv\) to get \(v = -\frac{1}{s+1}e^{-(s+1)t}\). This transforms the original integral into manageable components, allowing further simplification.
Exponential Functions
Exponential functions are mathematical expressions where the variable appears as the exponent. They form a core part of calculus and are especially relevant in the field of differential equations and the Laplace Transform.
- The general form is represented as \(e^{x}\), where \(e\) is Euler's constant, approximately equal to 2.71828.
- These functions exhibit rapid growth or decay depending on the sign of the exponent.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. They are essential tools in modeling dynamic systems where rates of change are studied, like population dynamics, motion, or electrical circuits.
- They appear in two main types: ordinary differential equations (ODEs) and partial differential equations (PDEs).
- These equations can often be solved using techniques like the Laplace Transform.