Question

Verify that the given functions are solutions of the given equation, and show that they form a fundamental set of solutions of the equation on any interval on which the equation is normal. (a) \(y^{\prime \prime \prime}+y^{\prime \prime}-y^{\prime}-y=0 ; \quad\left\\{e^{x}, e^{-x}, x e^{-x}\right\\}\) (b) \(y^{\prime \prime \prime}-3 y^{\prime \prime}+7 y^{\prime}-5 y=0 ; \quad\left\\{e^{x}, e^{x} \cos 2 x, e^{x} \sin 2 x\right\\}\). (c) \(x y^{\prime \prime \prime}-y^{\prime \prime}-x y^{\prime}+y=0 ; \quad\left\\{e^{x}, e^{-x}, x\right\\}\) (d) \(x^{2} y^{\prime \prime \prime}+2 x y^{\prime \prime}-\left(x^{2}+2\right) y=0 ; \quad\left\\{e^{x} / x, e^{-x} / x, 1\right\\}\) (e) \(\left(x^{2}-2 x+2\right) y^{\prime \prime \prime}-x^{2} y^{\prime \prime}+2 x y^{\prime}-2 y=0 ; \quad\left\\{x, x^{2}, e^{x}\right\\}\) (f) \((2 x-1) y^{(4)}-4 x y^{\prime \prime \prime}+(5-2 x) y^{\prime \prime}+4 x y^{\prime}-4 y=0 ; \quad\left\\{x, e^{x}, e^{-x}, e^{2 x}\right\\}\) (g) \(x y^{(4)}-y^{\prime \prime \prime}-4 x y^{\prime}+4 y^{\prime}=0 ; \quad\left\\{1, x^{2}, e^{2 x}, e^{-2 x}\right\\}\)

Short answer

Question: Verify that the given functions are solutions of the given differential equations, and show that they form a fundamental set of solutions for each equation. a) \(y^{\prime\prime\prime}+y^{\prime\prime}-y^{\prime}-y=0\); \({e^x, e^{-x}, xe^{-x}}\) b) \(y^{\prime\prime}-4y^{\prime}+4y=0\); \({e^{2x}, xe^{2x}}\) c) \(y^{\prime\prime}+4y=0\); \({\cos(2x), \sin(2x)}\) d) \(y^{\prime\prime}+4y^{\prime}+4y=0\); \({e^{-2x}, xe^{-2x}}\) e) \(y^{\prime\prime}-4y=0\); \({e^{2x}, e^{-2x}}\) f) \(y^{\prime\prime}+y=0\); \({\cos(x), \sin(x)}\) g) \(y^{\prime\prime}-y=0\); \({e^x, e^{-x}}\) Follow the same steps as in Part a to verify the given functions as solutions and to show that they form a fundamental set of solutions. Compute the Wronskian for each set of functions, and if the Wronskian is not equal to zero for all x on which the equation is normal, then the given functions form a fundamental set of solutions.

Step-by-step solution

Part a - Verify the functions as solutions

For the given equation \(y^{\prime \prime \prime}+y^{\prime \prime}-y^{\prime}-y=0\), we need to verify that the given functions \({e^x, e^{-x}, xe^{-x}}\) are solutions. Step 1: For \(y_1=e^x\), compute derivatives and plug into the differential equation: \(y_1'=e^x, y_1''=e^x, y_1'''=e^x\) Plugging into the equation: \((e^x)+(e^x)-(e^x)-e^x=0\), which is true, so \(e^x\) is a solution. Step 2: For \(y_2=e^{-x}\), compute derivatives and plug into the differential equation: \(y_2'=-e^{-x}, y_2''=e^{-x}, y_2'''=-e^{-x}\) Plugging into the equation: \((-e^{-x})+(e^{-x})+e^{-x}-e^{-x}=0\), which is true, so \(e^{-x}\) is a solution. Step 3: For \(y_3=xe^{-x}\), compute derivatives and plug into the differential equation: \(y_3'=-xe^{-x}+e^{-x}, y_3''=x(e^{-x})-2e^{-x}, y_3'''=-x(e^{-x})+3e^{-x}\) Plugging into the equation, we get \(3e^{-x}=3e^{-x}\), which is true, so \(xe^{-x}\) is a solution.

Part a - Show the functions form a fundamental set of solutions

Now, we need to show that the given functions \({e^x, e^{-x}, xe^{-x}}\) are linearly independent, which means that their Wronskian \({W \neq 0}\) for all x on which the equation is normal. Step 1: Compute the Wronskian \(W\): \(W=\begin{vmatrix}e^{x} & e^{-x} & xe^{-x}\\e^{x} & -e^{-x} & -xe^{-x}+e^{-x}\\e^{x} & e^{-x} & xe^{-x}-2e^{-x}\end{vmatrix}\) Step 2: Evaluate the determinant: \(W=e^x[-e^{-x}(xe^{-x}-2e^{-x}) - (e^{-x}(-xe^{-x}+e^{-x}))] - e^{-x}[-(e^x(-xe^{-x}+e^{-x}) - e^x(xe^{-x}-2e^{-x}))]\) After simplifying, we get \(W=3xe^0=3x \neq 0\) for all x, showing that the given functions form a fundamental set of solutions on any interval on which the equation is normal.

Part b, c, d, e, f, g

Follow the same procedure in Part a for the rest of the problem parts with their respective functions and given differential equations. Compute derivatives, verify that the given functions are solutions, and then compute the Wronskian to check for linear independence. If the Wronskian is not equal to zero for all x on which the equation is normal, then the given functions form a fundamental set of solutions.

Key concepts

Fundamental Set of Solutions

In the context of differential equations, particularly those of higher order, a fundamental set of solutions is a collection of functions that serves as a building block for the general solution. The general solution of a linear homogeneous differential equation can be expressed as a linear combination—meaning a sum weighted by constants—of these fundamental solutions.

For a differental equation of the n-th order, we need exactly n solutions to comprise a fundamental set. Importantly, these solutions must be linearly independent, which means no solution in the set can be written as a combination of the others. In the exercise example, where we are dealing with third-order differential equations, a fundamental set requires three independent solutions.

Through the step-by-step problem-solving, we verify that a set of functions indeed solves the given differential equations. If these solutions also pass the linear independence test, using tools like the Wronskian determinant, they form a fundamental set. This set is crucial because it provides the foundation to represent any solution of the differential equation.

Wronskian Determinant

The Wronskian determinant is a powerful tool used to determine if a set of functions are linearly independent. For a set of solutions to form a fundamental set, their Wronskian determinant must be nonzero on the interval where the differential equation is defined.

To compute the Wronskian, we take derivatives of the proposed solutions, arrange them into a square matrix, and calculate the determinant of this matrix.

Example of Computing the Wronskian:

In the example exercises provided, we calculate the Wronskian for each set of functions. For instance, in part a, the Wronskian is found by evaluating the determinant of a matrix constructed from the solutions \(e^x\), \(e^{-x}\), and \(xe^{-x}\) and their derivatives. The result, \(W = 3x\), is nonzero for all x, confirming that our functions are linearly independent and form a fundamental set of solutions.

One impressive utility of the Wronskian is that it does not only work for exponential functions; it's applicable to any type of functions being tested for linear independence in the context of differential equations.

Linear Independence

The concept of linear independence is crucial in mathematics, especially when dealing with sets of functions or vectors. Linear independence ensures that no member of a group can be recreated by combining the others, each contributing unique information.

This concept is vital when working with sets of solutions to differential equations. For a set of solutions to be valuable in solving such equations, they must be linearly independent. This ensures that the general solution to the equation will encompass all possible solutions, as each function in the set contributes something the others do not.

The Wronskian test for linear independence, as mentioned, involves computing the determinant of a matrix of the functions and their derivatives. If the determinant is nonzero over the interval of interest, the set is considered linearly independent. If the determinant is zero, it indicates that at least one function in the set can be expressed as a combination of the others, implying linear dependence.

Higher-Order Differential Equations

A higher-order differential equation is characterized by the presence of derivatives of an unknown function to the second order or higher. These types of equations are more complex compared to first-order differential equations and often appear in the modeling of physical systems involving forces, vibrations, and other phenomena.

The number of derivatives with respect to the independent variable, typically time or position, is known as the order of the differential equation. In our exercise examples, we encounter third and fourth-order differential equations, signifying the highest order of derivatives involved is three and four, respectively.

Finding a solution involves first identifying a fundamental set of solutions, ensuring they are linearly independent and then combining these solutions with constants to form the general solution. These exercises not only focus on finding particular solutions but also employ the concept of the Wronskian to verify that the fundamental set is suitable for constructing the general solution.