Problem 47 - Differential Equations and Linear Algebra

Question

We explore a relationship between determinants and solutions to a differential equation. The \(3 \times 3\) matrix consisting of solutions to a differential equation and their derivatives is called the Wronskian and, as we will see in later chapters, plays a pivotal role in the theory of differential equations. Verify that \(y_{1}(x)=\cos 2 x, y_{2}(x)=\sin 2 x,\) and \(y_{3}(x)=e^{x}\) are solutions to the differential equation \(y^{\prime \prime \prime}-y^{\prime \prime}+4 y^{\prime}-4 y=0\) 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

In summary, we verified that the given functions \(y_1(x) = \cos(2x), y_2(x) = \sin(2x),\) and \(y_3(x) = e^x\) are solutions to the given differential equation by finding their derivatives and substituting them into the equation. Next, we computed the determinant of the Wronskian matrix and showed that it is nonzero on any interval.

Step 1: Differentiate the given functions

First, find the first, second, and third derivatives of the given functions: \(\text{For } y_1(x) = \cos(2x):\) \[y_1'(x) = -2\sin(2x),\] \[y_1''(x) = -4\cos(2x),\] \[y_1'''(x) = 8\sin(2x).\] \(\text{For } y_2(x) = \sin(2x):\) \[y_2'(x) = 2\cos(2x),\] \[y_2''(x) = -4\sin(2x),\] \[y_2'''(x) = -8\cos(2x).\] \(\text{For } y_3(x) = e^{x}:\) \(y_3'(x) = e^{x},\) \(y_3''(x) = e^{x},\) \(y_3'''(x) = e^{x}.\)

Step 2: Verify the given functions as solutions

Now, we need to substitute the functions and their derivatives into the given differential equation to check if they satisfy the equation: \[y^{\prime\prime\prime} - y^{\prime\prime} + 4y^{\prime} - 4y = 0.\] For \(y_1(x)\): \[ 8\sin(2x) - (-4\cos(2x)) + 4(-2\sin(2x)) - 4\cos(2x) = 8\sin(2x) + 4\cos(2x) -8\sin(2x) - 4\cos(2x) =0.\] For \(y_2(x)\): \[- 8\cos(2x) - (-4\sin(2x)) + 4(2\cos(2x)) - 4\sin(2x) = -8\cos(2x) + 4\sin(2x) +8\cos(2x) -4\sin(2x) = 0.\] For \(y_3(x)\): \[e^x - e^x + 4e^x - 4e^x = 0.\] Therefore, \(y_1(x), y_2(x),\) and \(y_3(x)\) are indeed solutions to the given differential equation.

Step 3: Compute the determinant of the Wronskian matrix

Now, let's compute the determinant of the Wronskian matrix: \[W = \left|\begin{array}{ccc}\cos(2x) & \sin(2x) & e^x\\ -2\sin(2x) & 2\cos(2x) & e^x \\ -4\cos(2x) & -4\sin(2x) & e^x \\\end{array}\right|.\] Expanding the determinant along the first row: \[W =\cos(2x) \left|\begin{array}{cc}2\cos(2x) & e^x\\ -4\sin(2x) & e^x\end{array}\right| - \sin(2x) \left| \begin{array}{cc} -2\sin(2x) & e^x \\ -4\cos(2x) & e^x\end{array}\right| + e^x \left| \begin{array}{cc}-2\sin(2x) & 2\cos(2x) \\ -4\cos(2x) &-4\sin(2x) \\ \end{array}\right|.\] Computing the determinants of the submatrices: \[W = \cos(2x)(4e^x\cos(2x) + 4e^x\sin(2x)) - \sin(2x)(2e^x\sin(2x)+4e^x\cos(2x)) + e^x(16\sin^2(2x) + 16\cos^2(2x)).\] Factor \(4e^x\) from the sum and use the trigonometric identity \(\sin^2(x) + \cos^2(x) = 1\): \[W = 4e^x[\cos(2x)(\cos(2x) + \sin(2x)) - \sin(2x)(\frac{1}{2}\sin(2x)+\cos(2x)) + 4(\sin^2(2x) + \cos^2(2x))].\] Now, simplify the expression: \[W = 4e^x[\cos(2x)^2 + \cos(2x)\sin(2x) - \frac{1}{2}\sin^2(2x)-\sin(2x)\cos(2x)+4].\] Since \(e^x\) is always positive and the remaining expression is not zero, the Wronskian matrix is nonzero on any interval.

Key Concepts

Differential equations

Differential equations are mathematical equations that describe the relationships involving the rates at which quantities change. These tools are central in expressing the laws of nature and are widely used across science and engineering. They can range from simple first-order equations to complex partial differential equations with multiple variables.

Solving a differential equation typically means finding a function or a set of functions that satisfies the equation. There are various methods for solving differential equations, but one common approach is to use known solutions to construct the general solution of the equation. The particular solutions can sometimes be guessed through experience – as is often the case with trigonometric or exponential functions – but methods like separation of variables, integration factors, or utilizing a Wronskian determinant as seen in our example are often required.

Third-order linear differential equation

Third-order linear differential equations involve derivatives up to the third order. These equations have the general form:\[a_3(x)y''' + a_2(x)y'' + a_1(x)y' + a_0(x)y = g(x),\]where \(a_3(x)\), \(a_2(x)\), \(a_1(x)\), and \(a_0(x)\) are functions of \(x\), and \(g(x)\) is the source or forcing term. The equation presented in the exercise \(y''' - y'' + 4y' - 4y = 0\) is a homogeneous linear differential equation because \(g(x) = 0\).

To solve these equations, we typically look for solutions based on the nature of the coefficient functions. In the given problem, our solutions are in terms of trigonometric and exponential functions. Third-order linear differential equations can describe a variety of physical phenomena, such as the motion of a particle or the electric current in a circuit with inductance, resistance, and capacitance.

Trigonometric functions solutions

Trigonometric functions, like sine and cosine, are periodic and oscillatory, which makes them suitable for describing phenomena that repeat over time, such as waves and vibrations. When faced with a differential equation stemming from such contexts, solutions involving trigonometric functions are often expected.

In the textbook problem, two of the given functions, \(y_1(x) = \text{cos}(2x)\) and \(y_2(x) = \text{sin}(2x)\), are trigonometric. Their derivatives are also trigonometric functions, leading to a cyclical pattern that links back to the original functions. This property simplifies the process of checking if these functions satisfy the given third-order linear differential equation. Moreover, these functions' periodicity is particularly important when computing the Wronskian to show that the solutions are independent and span the solution space for the given differential equation.

Exponential functions solutions

Exponential functions are crucial when solving many types of differential equations, in particular when the equations involve growth or decay processes, such as populations, radioactive decay, or thermal cooling. Solutions involving exponential functions are characterized by their rate changes proportional to the value of the function itself.

In our exercise, the third function \(y_3(x) = e^x\) is an exponential function. Just like the trigonometric solutions, the derivatives of \(e^x\) lead to the same function, which again simplifies verification in the differential equation. The exponential function's contribution to the Wronskian determinant underscores the independence of the solutions and ensures that the determinant does not vanish, guaranteeing that the set of functions \{\cos(2x), \sin(2x), e^x\} constitutes a fundamental set of solutions for the differential equation.