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Magazine Chapter 2: Problem 5
. (a) Compute the Wronskian of four linearly independent solutions of the equation \(y^{(4)}+\) \(16 y=0 .\) (b) Compute that solution \(\phi\) of this equation which satisfies $$ \phi(0)=1, \quad \phi^{\prime}(0)=0, \quad \phi^{\prime \prime}(0)=0, \quad \phi^{\prime \prime \prime}(0)=0 $$
Short Answer
Expert verified
The Wronskian is non-zero, implying independence; the solution \( \phi(x) = \cos(2x) \).
Step by step solution
01
Identify the characteristic equation
For the differential equation \( y^{(4)} + 16y = 0 \), first identify the corresponding characteristic equation, which can be formed by replacing derivatives with powers of \( r \). This gives us \( r^4 + 16 = 0 \).
02
Solve the characteristic equation
To solve \( r^4 + 16 = 0 \), we can write it as \( r^4 = -16 \). Taking the fourth root of both sides gives us \( r = \pm 2i \). Since the roots are complex, they can be expressed in terms of their real and imaginary parts: \( r = \pm 2i \). However, to find all roots of \( -16 \), we also account for additional solutions: \( r = \pm 2 \cdot i \sqrt{-1} = \pm 2 \sqrt{-1} \cdot e^{i\frac{\pi}{2}} \) and \( r = \pm 2 \cdot i \cdot e^{-i\frac{\pi}{2}} \). Thus, the roots are \( 2i, -2i, 2i, -2i \).
03
Write the general solution of the differential equation
The general solution is a linear combination of the basic solutions given by oscillator functions stemming from our roots. Thus, we have:\[ y(x) = C_1 \cos(2x) + C_2 \sin(2x) + C_3x \cos(2x) + C_4x \sin(2x) \]
04
Calculate Wronskian of solutions
To compute the Wronskian of the solutions \( \{ \cos(2x), \sin(2x), x \cos(2x), x \sin(2x) \} \), form the Wronskian determinant. Evaluate it at some point (often easiest at \( x=0 \)): The matrix of derivatives:\[\begin{vmatrix}\cos(2x) & \sin(2x) & x\cos(2x) & x\sin(2x) \-2\sin(2x) & 2\cos(2x) & \cos(2x) - 2x\sin(2x) & \sin(2x) + 2x\cos(2x) \-4\cos(2x) & -4\sin(2x) & -4x\cos(2x) - 2\sin(2x) & -4x\sin(2x) + 2\cos(2x) \8\sin(2x) & -8\cos(2x) & -8\cos(2x) + 8x\sin(2x) & -8\sin(2x) - 8x\cos(2x)\end{vmatrix}\]Upon calculation, this determinant evaluates to a constant multiple, showing independence.
05
Find particular solution satisfying initial conditions
We are given the initial conditions at \( x=0 \) for \( \phi(x) \):\( \phi(0) = 1 \), \( \phi'(0) = 0 \), \( \phi''(0) = 0 \), \( \phi'''(0) = 0 \).Use the general solution and plug these conditions in to find the constants. Given:\[ \phi(x) = C_1 \cos(2x) + C_2 \sin(2x) + C_3 x \cos(2x) + C_4 x \sin(2x) \]Evaluate:\[ \phi(0) = C_1 = 1 \]\[ \phi'(0) = -2C_1 \sin(2\cdot 0) + 2C_2 \cos(2\cdot 0) + (C_3 \cos(2\cdot 0) - 2C_3 \cdot 0 \sin(2\cdot 0)) + (C_4 \sin(2\cdot 0) + 2C_4 \cdot 0 \cos(2\cdot 0)) = 0 \]This yields \( C_2 = 0 \), \( C_3 = 0 \), \( C_4 = 0 \). Hence, \( \phi(x) = \cos(2x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wronskian
The Wronskian is a mathematical tool used to determine whether a set of functions are linearly independent solutions to a differential equation. It is particularly useful in verifying the independence of solutions in systems of linear differential equations.
To compute the Wronskian for a set of functions, you use a determinant composed of those functions and their derivatives. For example, for the functions \( \cos(2x), \sin(2x), x\cos(2x), x\sin(2x) \), the Wronskian is constructed as follows: set up a matrix where each row consists of a function and its successive derivatives.
Calculate the determinant of this matrix at a particular point, often \( x = 0 \). If the Wronskian is non-zero, it confirms that the set of functions is linearly independent.
To compute the Wronskian for a set of functions, you use a determinant composed of those functions and their derivatives. For example, for the functions \( \cos(2x), \sin(2x), x\cos(2x), x\sin(2x) \), the Wronskian is constructed as follows: set up a matrix where each row consists of a function and its successive derivatives.
Calculate the determinant of this matrix at a particular point, often \( x = 0 \). If the Wronskian is non-zero, it confirms that the set of functions is linearly independent.
Characteristic Equation
The characteristic equation is a fundamental concept when solving linear differential equations. It helps find the roots that lead to the general solution of the differential equation.
For a given differential equation like \( y^{(4)} + 16y = 0 \), you form its characteristic equation by replacing each derivative with powers of \( r \). Thus, the characteristic equation becomes \( r^4 + 16 = 0 \).
By solving this polynomial equation, you find the roots that define the behavior of solutions. In this case, the complex roots \( 2i \) and \( -2i \) indicate oscillatory solutions like cosines and sines. These roots contribute directly to the general solution, providing the basis for further analysis.
For a given differential equation like \( y^{(4)} + 16y = 0 \), you form its characteristic equation by replacing each derivative with powers of \( r \). Thus, the characteristic equation becomes \( r^4 + 16 = 0 \).
By solving this polynomial equation, you find the roots that define the behavior of solutions. In this case, the complex roots \( 2i \) and \( -2i \) indicate oscillatory solutions like cosines and sines. These roots contribute directly to the general solution, providing the basis for further analysis.
Initial Value Problem
An initial value problem is a way to find a particular solution to a differential equation that satisfies given initial conditions. These conditions specify the value of the unknown function and possibly its derivatives at a particular point.
Consider the initial conditions \( \phi(0) = 1 \), \( \phi'(0) = 0 \), \( \phi''(0) = 0 \), and \( \phi'''(0) = 0 \) for the general solution \( \phi(x) = C_1 \cos(2x) + C_2 \sin(2x) + C_3x \cos(2x) + C_4x \sin(2x) \).
Here, you substitute these initial conditions into the general solution to determine the constants \( C_1, C_2, C_3, \) and \( C_4 \). By satisfying all conditions, you solve for specific values, arriving at a particular solution, such as \( \phi(x) = \cos(2x) \) for this example.
Consider the initial conditions \( \phi(0) = 1 \), \( \phi'(0) = 0 \), \( \phi''(0) = 0 \), and \( \phi'''(0) = 0 \) for the general solution \( \phi(x) = C_1 \cos(2x) + C_2 \sin(2x) + C_3x \cos(2x) + C_4x \sin(2x) \).
Here, you substitute these initial conditions into the general solution to determine the constants \( C_1, C_2, C_3, \) and \( C_4 \). By satisfying all conditions, you solve for specific values, arriving at a particular solution, such as \( \phi(x) = \cos(2x) \) for this example.
Linear Independence
Linear independence is a crucial concept when dealing with solutions to differential equations. It tells us whether a set of solutions provides a complete and non-redundant description of the solution space.
A set of functions is linearly independent if no function in the set can be expressed as a linear combination of the others. To test for independence, we use the Wronskian, as previously described.
If the Wronskian of a set of solutions is non-zero at some point, the solutions are linearly independent. This ensures each function contributes uniquely to the complete solution, allowing us to construct any solution from their linear combinations.
A set of functions is linearly independent if no function in the set can be expressed as a linear combination of the others. To test for independence, we use the Wronskian, as previously described.
If the Wronskian of a set of solutions is non-zero at some point, the solutions are linearly independent. This ensures each function contributes uniquely to the complete solution, allowing us to construct any solution from their linear combinations.
