Question

Use Abel’s formula (Problem 20) to find the Wronskian of a fundamental set of solutions of the given differential equation. $$ y^{\mathrm{iv}}+y=0 $$

Short answer

The Wronskian of a fundamental set of solutions of the given differential equation is \(-4\).

Step-by-step solution

Recall Abel's formula

Abel's formula states that for a linear homogeneous differential equation of the form: $$ L[y] = y^{(n)} + a_{n-1}(x)y^{(n-1)} + \dots + a_1(x)y' + a_0(x)y = 0, $$ where \(a_i(x)\) are continuous functions and \(y_1, y_2, \dots, y_n\) are fundamental solutions, the Wronskian \(W(y_1, y_2, \dots, y_n)\) can be given as follows: $$ W(y_1, y_2, \dots, y_n) = ce^{-\int a_{n-1}(x)dx}, $$ where \(c\) is a constant.

Find the fundamental solutions of the given equation

To find the fundamental solutions of the given equation, we first need to find the characteristic equation. The characteristic equation for the fourth-order equation is as follows: $$ r^4 + 1 = 0 $$ The solutions of the characteristic equation can be found by taking the roots of the equation. The characteristic roots are: \(r_1=i, r_2=-i, r_3=1, r_4=-1\). Now we can write the fundamental solutions based on the roots as: $$ y_1(x)=e^{ix}, y_2(x)=e^{-ix}, y_3(x)=e^x, y_4(x)=e^{-x} $$

Apply Abel's formula to find the Wronskian

From the given equation, the only coefficient present is \(a_{n-1}(x)=0\), because no third-order derivative term is present. As a result, Abel's formula is simplified to: $$ W(y_1, y_2, y_3, y_4) = ce^0 = c $$ Now we will calculate the Wronskian using the given solutions. The Wronskian for the four functions is: $$ W = \begin{vmatrix} y_1 & y_2 & y_3 & y_4 \\ y_1' & y_2' & y_3' & y_4' \\ y_1'' & y_2'' & y_3'' & y_4'' \\ y_1^{(3)} & y_2^{(3)} & y_3^{(3)} & y_4^{(3)} \end{vmatrix} $$ By taking the derivatives of the fundamental solutions, we can fill the matrix and compute the determinant. The result will be the Wronskian, which is a constant.

Calculate the Wronskian

After calculating the derivatives and filling the matrix, we find the determinant. To make it easier, let's first take the derivatives: \(y_1'(x)=ie^{ix}, y_2'(x)=-ie^{-ix}, y_3'(x)=e^x, y_4'(x)=-e^{-x}\) \(y_1''(x)=-e^{ix}, y_2''(x)=e^{-ix}, y_3''(x)=e^x, y_4''(x)=e^{-x}\) \(y_1^{(3)}(x)=-ie^{ix}, y_2^{(3)}(x)=ie^{-ix}, y_3^{(3)}(x)=e^x, y_4^{(3)}(x)=-e^{-x}\) Now, substitute these into the Wronskian matrix and calculate the determinant: $$ W = \begin{vmatrix} e^{ix} & e^{-ix} & e^x & e^{-x} \\ ie^{ix} & -ie^{-ix} & e^x & -e^{-x}\\ -e^{ix} & e^{-ix} & e^x & e^{-x} \\ -ie^{ix} & ie^{-ix} & e^x & -e^{-x} \end{vmatrix} $$ Calculating the determinant, we find that the Wronskian is equal to \(-4e^{ix}e^{-ix}=-4\). So, we have: $$ W(y_1, y_2, y_3, y_4) = -4 $$ Thus, the Wronksian of a fundamental set of solutions of the given differential equation is \(-4\).

Key concepts

Abel's formula

Abel's formula is a handy tool for understanding Wronskians in linear homogeneous differential equations. It provides a way to compute the Wronskian of a set of fundamental solutions without directly evaluating the determinant of a large matrix, which can be cumbersome and complex. The formula

• is mainly used for nth order linear homogeneous differential equations
• states that the Wronskian, W, is expressed as: \( W(y_1, y_2, \ldots, y_n) = ce^{-\int a_{n-1}(x)dx} \)

The constant \( c \) arises from initial conditions, and the integral term, \( a_{n-1}(x) \), is from the coefficient of the \( y^{(n-1)} \) term of the differential equation.
In our exercise, since the coefficient \( a_{n-1}(x) \) is zero (due to missing third order term), the Wronskian simplifies to be a constant.

fourth-order differential equation

Fourth-order differential equations are equations that involve fourth derivatives of a function. The highest order derivative term—\( y^{\text{iv}} \) in this context—determines the order of the equation. These equations are common in various physical systems that model phenomena like beam deflection and oscillations.
In the given problem, the differential equation is \( y^{\text{iv}} + y = 0 \). This structure means the solution involves understanding how both the fourth derivative and the original function, \( y \), interplay to form solutions.
When solving said differential equations, a characteristic equation is often formulated to determine solutions that reflect the behavior of the system described.

characteristic equation

The characteristic equation is a central concept when solving linear differential equations with constant coefficients. This polynomial equation arises from substituting trial solutions like \( y = e^{rx} \) into the differential equation, resulting in finding possible solutions (roots).
In the exercise, we start with the fourth-order differential equation \( y^{\text{iv}} + y = 0 \). By substituting \( y = e^{rx} \), we obtain the characteristic equation \( r^4 + 1 = 0 \). Solving this equation gives roots, which in this case are \( r = i, -i, 1, -1 \).
These roots guide us to the fundamental solutions, consisting of complex exponentials \( e^{ix}, e^{-ix} \) and real exponentials \( e^{x}, e^{-x} \), which form a complete set for the differential equation.

fundamental solutions

Fundamental solutions are the building blocks for solving linear differential equations. For a fourth-order equation, we need four fundamental solutions, derived from the roots of the characteristic equation.
For the differential equation \( y^{\text{iv}} + y = 0 \), we find the characteristic roots: \( r_1 = i, r_2 = -i, r_3 = 1, r_4 = -1 \). These roots lead to fundamental solutions:

• \( y_1(x) = e^{ix} \)
• \( y_2(x) = e^{-ix} \)
• \( y_3(x) = e^x \)
• \( y_4(x) = e^{-x} \)

Combining these solutions, we can form a general solution for the differential equation, representing any solution as a linear combination of the fundamental solutions: \( y = c_1 y_1 + c_2 y_2 + c_3 y_3 + c_4 y_4 \). This linear combination allows for capturing all possible behaviors described by the differential equation.