Question

Take it as given that $\text{Missing left or extra right}$ and $x\cos kx,x\sin kx$ are fundamental sets of solutions of $$x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y-k^2{x}^{2}y=0$$ and $$x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y+k^2{x}^{2}y=0$$ respectively. Solve the eigenvalue problem for $$x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y+\lambda x^{2}y=0,y(1)=0,y(2)=0$$

Short answer

#Answer# The eigenvalue problem can be transformed into two simpler differential equations using the provided fundamental sets of solutions. The boundary conditions must then be applied to these equations to find the eigenvalues (k values) that satisfy the conditions. The transformed eigenvalue problem can be written in matrix form and its determinant must be calculated and set to zero to find valid k values.

Step-by-step solution

Understand the given information

We are given fundamental sets of solutions for two differential equations and need to solve an eigenvalue problem for a third equation, with two boundary conditions. Our job is to use the information given to solve this new equation.

Rewrite the eigenvalue equation in a known form

We have the eigenvalue problem: $$x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y+\lambda x^{2}y=0$$ Comparing this equation with the previously analyzed ones, we see that replacing $\lambda$ with $-k^2$ or $+k^2$ gives us the same equations from the start. Therefore, we have two cases to examine: 1. $\lambda =-k^2$: $x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y-k^2{x}^{2}y=0$ 2. $\lambda =k^2$: $x^2{y}^{\prime \prime }-2x{y}^{\prime }+2y+k^2{x}^{2}y=0$

Find solutions for both cases

Now we have two cases and for each case we have a fundamental set of solutions. 1. For $\lambda =-k^2$, we have the fundamental set of solutions: $\{x{e}^{kx},x{e}^{-kx}\}$ 2. For $\lambda =k^2$, we have the fundamental set of solutions: $\{x\cos kx,x\sin kx\}$

Apply the boundary conditions

We need to find the solutions that satisfy the given boundary conditions $y(1)=0$ and $y(2)=0$. For the first case with the fundamental solutions $\{x{e}^{kx},x{e}^{-kx}\}$: $$y(1)=A{e}^k+A{e}^{-k}=0$$ $$y(2)=2A{e}^{2k}+2A{e}^{-2k}=0$$ Since we want non-trivial solutions ($A\ne 0$), we need the functions inside the sum to cancel each other (hyperbolic sine function sinh). However, it fails for the second boundary condition in this case, so this set of solutions doesn't satisfy the conditions. Now let's check the second case with the fundamental solutions $\{x\cos kx,x\sin kx\}$: $$y(1)=A\cos k+\sin k=0$$ $$y(2)=2A\cos 2k+2\sin 2k=0$$ Dividing the second equation by $2$, we have: $$A\cos 2k+\sin 2k=0$$ These equations form a system of homogeneous linear equations for $A$ and $k$. To have non-trivial solutions, the determinant of the matrix must be zero.

Find the eigenvalues (k values) satisfying the conditions

We have the following matrix for the coefficients of A and k: $$\left(\begin{array}{cc}\cos k& 1\\ \cos 2k& \sin 2k\end{array}\right)$$ Calculating the determinant and setting it to zero: $$det\left(\begin{array}{cc}\cos k& 1\\ \cos 2k& \sin 2k\end{array}\right)=(\cos k)(\sin 2k)-(\cos 2k)(1)=0$$ Solving this equation, we can find the values of $k$ that satisfy the boundary conditions. These values are the eigenvalues.

Conclusion

To summarize, we used the provided information on fundamental sets of solutions to rewrite the given eigenvalue problem into two simpler, known-structure differential equations. We then applied the boundary conditions to these equations to find the eigenvalues (k values) that satisfy the conditions.

Key concepts

Boundary Conditions

Boundary conditions are constraints necessary for solving differential equations, particularly in an eigenvalue problem. They define the values that a solution must take on the boundaries of the domain. In this exercise, we have two boundary conditions: $y(1)=0$ and $y(2)=0$.

These values mean that the solution to the differential equation must meet these conditions exactly at the endpoints $x=1$ and $x=2$.

By setting these criteria, boundary conditions help accurately determine the specific eigenvalues for the problem. They ensure that the mathematical model represented by the differential equation aligns closely with real-world or theoretical constraints. Boundary conditions thus play a critical role in defining the behavior and uniqueness of solutions to differential equations.

Fundamental Solutions

Fundamental solutions are essential building blocks for solving differential equations. They represent a set of solutions that can generate the general solution through linear combinations.

In this exercise, the fundamental solutions provided are specific sets corresponding to different values of $\lambda$ when equated to $-k^2$ and $k^2$.

The fundamental sets $\{x{e}^{kx},x{e}^{-kx}\}$ and $\{x\cos kx,x\sin kx\}$ offer the solutions under the conditions where $\lambda =-k^2$ and $\lambda =k^2$, respectively.

These solutions form the basis from which all particular solutions to the differential equation can be constructed. Each fundamental solution represents a different type of behavior (exponential, oscillatory) that the original equation can exhibit, determined by the nature of the eigenvalue $\lambda$.

By applying the boundary conditions to these fundamental solutions, we find the specific combination that satisfies the original differential equation.

Homogeneous Linear Equations

Homogeneous linear equations are a type of differential equation where the right-hand side is zero. This makes the equation uniform in its structure, meaning all terms are dependent on the unknown function and its derivatives.

In this exercise, the given differential equations take the form $x^2{y}^{″}-2x{y}^{\prime }+2y\pm k^2{x}^{2}y=0$, indicating a homogeneous equation.

The solutions to these equations, particularly in the context of eigenvalue problems, depend heavily on the boundary conditions provided. As they are homogeneous, any linear combination of solutions that satisfy the boundary conditions remains a solution.

Solving homogeneous linear differential equations involves finding values of parameters (like $\lambda$ here) that enable non-trivial solutions. These solutions must not only satisfy the equation but also meet the additional criteria from boundary conditions.

Typically, these equations require techniques such as finding the determinant of a matrix, ensuring it equals zero, to find permissible eigenvalues without trivial solutions.

Differential Equations

Differential equations involve mathematical expressions relating functions and their derivatives. They express how a quantity changes concerning others. In physics and engineering, these equations often model various phenomena.

In this specific problem, the differential equation $x^2{y}^{″}-2x{y}^{\prime }+2y+\lambda x^{2}y=0$ represents a type of Bessel equation. This equation form is typical in problems involving wave propagation or heat conduction.

Solving such equations generally requires the initial step of identifying a fundamental set of solutions. These solutions help to form a variety of possible outputs of the equation under given constraints.

Eigenvalue problems, like the one presented, involve finding special values (eigenvalues) that allow the differential equation a solution satisfying specified boundary conditions.

Differential equations and their solutions are pivotal in translating physical concepts into a mathematical framework, where real-world phenomena can be analyzed and understood through the interplay of boundary conditions and parameters.