Chapter 8: Problem 4
Find \(L y\) for the given differential operator if \((a) y(x)=2 e^{3 x},\) (b) \(y(x)=3 \ln x,(c) y(x)=\) \(2 e^{3 x}+3 \ln x\). $$L=D^{3}-D+4$$
Short Answer
Expert verified
For the given functions and differential operator, we have:
(a) For \(y(x) = 2e^{3x}\), \(Ly = 56e^{3x}\)
(b) For \(y(x) = 3\ln x\), \(Ly = \frac{6}{x^3} - \frac{3}{x} + 12\ln(x)\)
(c) For \(y(x) = 2e^{3x} + 3\ln x\), \(Ly = 56e^{3x} + \frac{6}{x^3} - \frac{3}{x} + 12\ln(x)\)
Step by step solution
01
Apply the differential operator to y(x) = 2e^(3x)
We have \(L = D^3 - D + 4\) and \(y(x) = 2e^{3x}\). To find \(Ly\), we apply the differential operator to \(y(x)\):
\(Ly = (D^3 - D + 4)y(x) = (D^3 - D + 4)(2e^{3x})\)
Now, we will differentiate the function three times, and then subtract the first derivative and finally add four times the original function.
02
Differentiate y(x) = 2e^(3x) three times
Find the first, second, and third derivatives of \(y(x)=2e^{3x}\):
\(y'(x)=\frac{d}{dx}(2e^{3x})=6e^{3x}\)
\(y''(x)=\frac{d}{dx}(6e^{3x})=18e^{3x}\)
\(y'''(x)=\frac{d}{dx}(18e^{3x})=54e^{3x}\)
03
Plug the derivatives back into the differential operator
Now that we have the derivatives, we can plug them back into the expression for \(Ly\):
\(Ly = (D^3 - D + 4)y(x) = y'''(x) - y'(x) + 4y(x) = 54e^{3x} - 6e^{3x} + 8e^{3x} = 56e^{3x}\)
So for function (a) \(y(x) = 2e^{3x},\) we have:
\(Ly = 56e^{3x}\)
Now, let's repeat the process for the other two functions.
04
Apply the differential operator to y(x) = 3ln(x)
We have \(L = D^3 - D + 4\) and \(y(x) = 3\ln x\). We apply the differential operator to \(y(x)\):
\(Ly = (D^3 - D + 4)y(x) = (D^3 - D + 4)(3\ln x)\)
Now, we will differentiate the function three times, and then subtract the first derivative and finally add four times the original function.
05
Differentiate y(x) = 3ln(x) three times
Find the first, second, and third derivatives of \(y(x) = 3\ln x\):
\(y'(x)=\frac{d}{dx}(3\ln x) = \frac{3}{x}\)
\(y''(x)=\frac{d}{dx}(\frac{3}{x}) = -\frac{3}{x^2}\)
\(y'''(x)=\frac{d}{dx}(-\frac{3}{x^2}) = \frac{6}{x^3}\)
06
Plug the derivatives back into the differential operator
Now that we have the derivatives, we can plug them back into the expression for \(Ly\):
\(Ly = (D^3 - D + 4)y(x) = y'''(x) - y'(x) + 4y(x) = \frac{6}{x^3} - \frac{3}{x} + 12\ln(x) = \frac{6}{x^3} - \frac{3}{x} + 12\ln(x)\)
So for function (b) \(y(x) = 3\ln x,\) we have:
\(Ly = \frac{6}{x^3} - \frac{3}{x} + 12\ln(x)\)
07
Apply the differential operator to y(x) = 2e^(3x) + 3ln(x)
We have \(L = D^3 - D + 4\) and \(y(x) = 2e^{3x} + 3\ln x\). We apply the differential operator to \(y(x)\):
\(Ly = (D^3 - D + 4)y(x) = (D^3 - D + 4)(2e^{3x} + 3\ln x)\)
Since the differential operator behaves as a linear operator, we can apply it separately to each term:
\(Ly = (D^3 - D + 4)(2e^{3x}) + (D^3 - D + 4)(3\ln x)\)
From Steps 1-6, we already found that \(L(2e^{3x}) = 56e^{3x}\) and \(L(3\ln x) = \frac{6}{x^3} - \frac{3}{x} + 12\ln(x)\). Therefore:
\(Ly = 56e^{3x} + \frac{6}{x^3} - \frac{3}{x} + 12\ln(x)\)
So for function (c) \(y(x) = 2e^{3x} + 3\ln x,\) we have:
\(Ly = 56e^{3x} + \frac{6}{x^3} - \frac{3}{x} + 12\ln(x)\)
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.
Differentiation
Differentiation is the process of finding the derivative of a function, which is a measure of how the function changes as its input changes. It is a core concept in calculus and has many applications in mathematics and the sciences.
- The derivative of a function at a point is the slope of the tangent to the graph of the function at that point.
- The notation \( \frac{d}{dx} \) is commonly used to denote the differentiation operation with respect to \( x \).
- The first derivative \( y'(x) \) is found by applying \( \frac{d}{dx}(2e^{3x}) = 6e^{3x} \).
- The second derivative, \( y''(x) \), is then \( \frac{d}{dx}(6e^{3x}) = 18e^{3x} \).
- The third derivative, \( y'''(x) \), results in \( \frac{d}{dx}(18e^{3x}) = 54e^{3x} \).
Exponential Functions
Exponential functions are mathematical functions of the form \( f(x) = a e^{bx} \), where \( e \) is the base of natural logarithms (approximately equal to 2.71828), and \( a \) and \( b \) are constants. These functions are notable for their rapid growth and unique properties.
- An exponential function grows or decays at a rate proportional to its current value.
- The derivative of an exponential function retains the same exponential structure, making differentiation straightforward.
- In our exercise, the function \( y(x) = 2e^{3x} \) follows this pattern, with a constant factor \( a = 2 \) and an exponent coefficient \( b = 3 \).
- The result is another exponential function, multiplied by the derivative of the exponent \( 3 \), because \( \frac{d}{dx}e^{3x} = 3e^{3x} \).
- Consequently, operations such as differentiation maintain the function's essential form and structure.
Logarithmic Functions
Logarithmic functions, particularly the natural logarithm \( \,\ln x \, \), are inverse operations to exponentiation. They are defined for positive \( x \) and are fundamental in various branches of mathematics and science.
- The logarithmic function \( y(x) = \,\ln x \, \), represents the power to which the base \( e \) must be raised to yield \( x \).
- One of the primary characteristics of logarithms is that their growth rate reduces as \( x \) increases, leading them to be considered slowly increasing functions.
- The first derivative \( y'(x) \) of \( 3\ln x \) is \( \frac{3}{x} \), which inversely relates to the original function's input.
- Continuing to differentiate yields the second derivative \( y''(x) = -\frac{3}{x^2} \), and further to the third derivative \( y'''(x) = \frac{6}{x^3} \).