Problem 3 - Differential Equations and Linear Algebra

Question

Find \(L y\) for the given differential operator if \((a) y(x)=2 e^{3 x},\) (b) \(y(x)=3 \ln x,(c) y(x)=\) \(2 e^{3 x}+3 \ln x\). $$L=D^{3}-2 x D^{2}$$

Step-by-Step Solution

Answer

In short, the results of applying the differential operator \(L = D^3 - 2x D^2\) to the given functions are: (a) For \(y(x) = 2e^{3x}\), we have \(Ly(x) = 54e^{3x} - 36xe^{3x}\). (b) For \(y(x) = 3\ln{x}\), we have \(Ly(x) = \frac{6}{x^3} + 6\). (c) For \(y(x) = 2e^{3x} + 3\ln{x}\), we have \(Ly(x) = 54e^{3x} - 36xe^{3x} + \frac{6}{x^3} + 6\).

Step 1: Applying the differential operator to \(y(x) = 2e^{3x}\)

First, we want to find the result of applying the differential operator \(L = D^3 - 2x D^2\) to the function \(y(x) = 2e^{3x}\). Find the first derivative: \(y'(x) = \frac{d}{dx}(2e^{3x}) = 6e^{3x}\) Find the second derivative: \(y''(x) = \frac{d^2}{dx^2}(2e^{3x}) = 18e^{3x}\) Find the third derivative: \(y'''(x) = \frac{d^3}{dx^3}(2e^{3x}) = 54e^{3x}\) Next, we apply the operator \(L = D^3 - 2x D^2\): \(Ly(x) = (54e^{3x}) - 2x(18e^{3x})\) Finally, we simplify the expression: \(Ly(x) = 54e^{3x} - 36xe^{3x}\)

Step 2: Applying the differential operator to \(y(x) = 3\ln{x}\)

Now, we want to apply the differential operator \(L = D^3 - 2x D^2\) to the function \(y(x) = 3\ln{x}\). Find the first derivative: \(y'(x) = \frac{d}{dx}(3\ln{x}) = \frac{3}{x}\) Find the second derivative: \(y''(x) = \frac{d^2}{dx^2}(3\ln{x}) = -\frac{3}{x^2}\) Find the third derivative: \(y'''(x) = \frac{d^3}{dx^3}(3\ln{x}) = \frac{6}{x^3}\) Next, we apply the operator \(L = D^3 - 2x D^2\): \(Ly(x) = \left(\frac{6}{x^3}\right) - 2x\left(-\frac{3}{x^2}\right)\) Finally, simplify the expression: \(Ly(x) = \frac{6}{x^3} + 6\)

Step 3: Applying the differential operator to \(y(x) = 2e^{3x} + 3\ln{x}\)

Lastly, we want to apply the differential operator \(L = D^3 - 2x D^2\) to the function \(y(x) = 2e^{3x} + 3\ln{x}\). There is a key property of differential operators: \(L[y(x)+z(x)] = Ly(x) + Lz(x)\). Therefore, we can apply the operator to each function separately and then add the results. Applying the operator to \(y(x) = 2e^{3x}\) is already done in Step 1, and the result is, \(Ly(x) = 54e^{3x} - 36xe^{3x}\). Applying the operator to \(y(x) = 3\ln{x}\) is already done in Step 2, and the result is, \(Ly(x) = \frac{6}{x^3} + 6\). Now, we add the results for both parts: \(Ly(x) = (54e^{3x} - 36xe^{3x}) + \left(\frac{6}{x^3} + 6\right)\) Finally, we write the final result: \(Ly(x) = 54e^{3x} - 36xe^{3x} + \frac{6}{x^3} + 6\)

Key Concepts

Linear Differential Equations

Linear differential equations play an important role in mathematics and science, serving as a tool to describe a variety of real-world situations. A `linear differential equation` is one in which the dependent variable and all its derivatives appear to the first power and there are no products of these variables.
This is crucial because it distinguishes linear differential equations from non-linear ones, which can be more difficult to solve.
You must understand the basic structure of a linear differential equation because it often allows the use of powerful mathematical techniques like superposition. Superposition implies that if two functions are solutions of a linear differential equation, their sum is also a solution.

• One common form of a linear differential equation is: \(a_n D^n y + a_{n-1} D^{n-1} y + \ldots + a_1 D y + a_0 y = f(x)\), where \(D\) represents differentiation.
• Linearity means the coefficients \(a_n, a_{n-1}, \ldots, a_0\) are functions of the independent variable \(x\) only.
• Solutions often involve functions like exponentials and polynomials, making them versatile.

For example, in our case, the differential operator is \(L = D^3 - 2xD^2\), and our task was to apply it to different functions to check for solutions.

Exponential Functions

Exponential functions are a fundamental part of mathematics, defining a class of functions with rapid growth or decay properties. They can be generally represented as \(y(x) = ae^{bx}\), where \(a\) and \(b\) are constants. The unique feature of an exponential function is that the rate of change is proportional to the value of the function itself.
This makes exponential functions highly applicable in contexts like population growth, radioactive decay, and interest calculations.

• The base \(e\) is an irrational number approximately equal to 2.71828, known as Euler's number, and is often used because of its natural properties.
• Exponentials are notable for their simple differentiation properties: the derivative of \(e^{bx}\) is \(be^{bx}\).
• In the differential operator context, exponential functions can lead to distinctive results when derivatives are calculated, as shown through the transformation \(y(x) = 2e^{3x}\).

Understanding these characteristics is essential for handling various scientific and engineering problems.

Logarithmic Functions

Logarithmic functions are closely related to exponential functions and are defined as the inverse. The natural logarithm \(\ln x\) is widely used, both in pure and applied mathematics, due to its simplifying properties in calculus and algebra.
Logarithmic functions are particularly beneficial when dealing with problems involving multiplicative processes.

• The derivative of \(\ln x\) is \(1/x\), which influences how transformations behave under calculus operations.
• When differentiating twice, you obtain \(-1/x^2\), leading to transformations that adjust ratios instead of absolute differences.
• The changes contribute to understanding dynamic systems, such as in algorithms complexity or signal processing.

In the context of a differential operator application, logarithmic functions, such as \(y(x) = 3\ln x\), undergo transformations easily, letting us resolve complex systems by breaking down tasks into manageable steps.