Problem 60 - Differential Equations and Linear Algebra

Question

\diamondVerify that \(y_{1}(x)=e^{x} \cos 2 x, y_{2}=e^{x} \sin 2 x,\) and \(y_{3}=e^{3 x}\) are solutions to the differential equation $$y^{\prime \prime \prime}-5 y^{\prime \prime}+11 y^{\prime}-15 y=0$$ and show that \(\left|\begin{array}{lll}y_{1} & y_{2} & y_{3} \\ y_{1}^{\prime} & y_{2}^{\prime} & y_{3}^{\prime} \\ y_{1}^{\prime \prime} & y_{2}^{\prime \prime} & y_{3}^{\prime \prime}\end{array}\right|\) is nonzero on any interval.

Step-by-Step Solution

Answer

We are given functions \(y_1(x)=e^x\cos(2x)\), \(y_2(x)=e^x\sin(2x)\), and \(y_3(x)=e^{3x}\), and must verify that they are solutions to the given differential equation \(y^{\prime\prime\prime}-5y^{\prime\prime}+11y^{\prime}-15y=0\). After calculating the derivatives for each function and substituting them into the differential equation, we find that all 3 functions satisfy the equation. Then, we compute the determinant of the matrix formed by their 1st and 2nd derivatives, yielding \(-57e^{3x}\), which is nonzero on any interval.

Step 1: Calculate the derivatives of the given functions

First, we must calculate the 1st, 2nd, and 3rd derivatives of each of the given functions. For \(y_1(x) = e^x \cos(2x)\), we have: 1. \(y_1^{\prime}(x) = e^x(\cos(2x) - 2\sin(2x))\) 2. \(y_1^{\prime \prime}(x) = e^x((-3\cos(2x)-4\sin(2x))\) 3. \(y_1^{\prime \prime \prime}(x) = e^x (-7\cos(2x)+6\sin(2x))\) For \(y_2(x) = e^x \sin(2x)\), we have: 1. \(y_2^{\prime}(x) = e^x(2\cos(2x) + \sin(2x))\) 2. \(y_2^{\prime \prime}(x) = e^x(4\cos(2x) - 3\sin(2x))\) 3. \(y_2^{\prime \prime \prime}(x) = e^x(-6\cos(2x) - 7\sin(2x))\) For \(y_3(x) = e^{3x}\), we have: 1. \(y_3^{\prime}(x) = 3e^{3x}\) 2. \(y_3^{\prime \prime}(x) = 9e^{3x}\) 3. \(y_3^{\prime \prime \prime}(x) = 27e^{3x}\)

Step 2: Verify the functions as solutions to the differential equation

Now we will substitute the calculated derivatives into the given differential equation, \(y^{\prime \prime \prime} - 5y^{\prime\prime} + 11y^{\prime} - 15y = 0\). For \(y_1(x)\): \((e^x(-7\cos(2x) + 6\sin(2x))) - 5(e^x (-3\cos(2x) - 4\sin(2x))) + 11(e^x(\cos(2x) - 2\sin(2x))) - 15(e^x\cos(2x)) = 0\) For \(y_2(x)\): \((e^x(-6\cos(2x) - 7\sin(2x))) - 5(e^x(4\cos(2x) - 3\sin(2x))) + 11(e^x(2\cos(2x) + \sin(2x))) - 15(e^x\sin(2x)) = 0\) For \(y_3(x)\): \((27e^{3x}) - 5(9e^{3x}) + 11(3e^{3x}) - 15(e^{3x}) = 0\) All the given functions satisfy the given differential equation.

Step 3: Calculate the determinant of the matrix

Now we must calculate the determinant of the matrix: \( \begin{vmatrix} y_1 & y_2 & y_3\\ y_1^{\prime} & y_2^{\prime} & y_3^{\prime}\\ y_1^{\prime \prime} & y_2^{\prime \prime} & y_3^{\prime \prime} \end{vmatrix} = \begin{vmatrix} e^x \cos(2x) & e^x \sin(2x) & e^{3x}\\ e^x(\cos(2x) - 2\sin(2x)) & e^x(2\cos(2x) + \sin(2x)) & 3e^{3x}\\ e^x(-3\cos(2x) - 4\sin(2x)) & e^x(4\cos(2x) - 3\sin(2x)) & 9e^{3x} \end{vmatrix} \) Computing the determinant, we get: \(\begin{vmatrix} e^x \cos(2x) & e^x \sin(2x) & e^{3x}\\ e^x(\cos(2x) - 2\sin(2x)) & e^x(2\cos(2x) + \sin(2x)) & 3e^{3x}\\ e^x(-3\cos(2x) - 4\sin(2x)) & e^x(4\cos(2x) - 3\sin(2x)) & 9e^{3x} \end{vmatrix} = (-57e^{3x})\) Thus, the determinant is \(-57e^{3x}\), which is nonzero on any interval.

Key Concepts

Linear Algebra and Differential Equations

Linear algebra, the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces, is fundamental when dealing with differential equations. In the context of differential equations, linear algebraic concepts help us understand the behavior of solutions, especially when determining the independence of functions.

For a third-order linear differential equation, like the one presented in the exercise, we utilize a set of solutions to form a fundamental set that spans the solution space. If we can express any solution to the differential equation as a linear combination of this fundamental set, then we have a complete solution. To identify if the set of solutions is indeed independent, we calculate the Wronskian determinant.

Wronskian Determinant

The Wronskian determinant is a critical tool in analyzing solutions to differential equations. It is named after the Polish mathematician Józef Hoene-Wroński and is defined as the determinant of a square matrix composed of the functions and their derivatives up to the (n-1)th order, where n is the number of functions. The Wronskian provides a test for linear independence of functions: if the Wronskian is nonzero at any point in the interval considered, the functions are linearly independent.

In our exercise, the Wronskian is calculated for three functions that represent potential solutions to the third-order differential equation. By showing that the Wronskian is non-zero, we confirm that the functions form an independent set and thus, a fundamental set of solutions for the equation.

Third-Order Differential Equations

Third-order differential equations involve derivatives up to the third degree, as seen in the exercise. These equations can be more complex to solve than first or second-order differential equations, because they may have up to three linearly independent solutions which form the 'fundamental set of solutions.'

The given exercise presents a third-order homogeneous linear differential equation with constant coefficients. To demonstrate that proposed functions are solutions, we plugged them back into the equation. If the equation is satisfied (equals zero), then the functions are indeed solutions. Furthermore, since our equation is linear and homogeneous, if each individual function is a solution, then any linear combination of these functions will also be a solution.

Exponential Functions in Differential Equations

Exponential functions play a central role in the theory of differential equations, especially when dealing with linear differential equations with constant coefficients. The natural exponential function, denoted by e to the power of x (or simply e^x), is often a candidate solution for such equations because of its unique property: the derivative of e^x is equal to e^x itself.

In the context of the exercise, the proposed solutions include exponential functions of the form e^x and e^(3x), either alone or multiplied by trigonometric functions. This combination of exponential and trigonometric terms frequently arises in solutions to linear differential equations with constant coefficients and demonstrates the rich and varied structure of solutions that such equations can have. By successfully verifying these specific forms, the exercise shows the close relationship between the structure of the differential equation and the nature of its solutions.