Question

The differential equations in Problems 19 and 20 differ from those in previous problems in that the parameter \(\lambda\) multiplies the \(y^{\prime}\) term as well as the \(y\) term. In each of these problems determine the real eigenvalues and the corresponding eigenfunctions. $$ \begin{array}{l}{y^{\prime \prime}+y^{\prime}+\lambda\left(y^{\prime}+y\right)=0} \\ {y^{\prime}(0)=0, \quad y(1)=0}\end{array} $$

Short answer

Question: Determine the real eigenvalues and the corresponding eigenfunctions of the given differential equation with boundary conditions: Differential equation: \(y''+y'+\lambda(y'+y)=0\) Boundary conditions: \(y'(0)=0\), \(y(1)=0\) Answer: The eigenvalues and eigenfunctions for the given differential equation with boundary conditions can be found using numerical methods to solve the homogeneous system of linear equations. The eigenvalues \(\lambda_n\) and the corresponding eigenfunctions are in the form: $$y_n(x) = C_1e^{m_1x} + C_2e^{m_2x}$$ where n denotes the index of the eigenvalue-eigenfunction pair, with \(m_1\) and \(m_2\) solving the quadratic equation: $$m^2+(1+\lambda_n)m+\lambda_n = 0$$

Step-by-step solution

Rewrite the given differential equation

Rewrite the given differential equation as: $$y''+y'+\lambda(y'+y)=0$$ It can be grouped as: $$y''+(1+\lambda)y'+\lambda y = 0$$

Solve the differential equation

The above equation is a second-order linear homogeneous differential equation with constant coefficients. We assume a solution of the form: $$y(x) = e^{mx}$$ Now we compute the first and second derivatives: $$y'(x) = me^{mx}, \hspace{5mm} y''(x) = m^2e^{mx}$$ Substitute the assumed solution and its derivatives back into the differential equation: $$m^2e^{mx}+(1+\lambda)me^{mx}+\lambda e^{mx}=0$$ Since \(e^{mx}\neq 0\), we can simplify the equation by dividing through by \(e^{mx}\): $$m^2+(1+\lambda)m+\lambda = 0$$

Find the values of m

Now we solve the quadratic equation for the values of m: $$m^2+(1+\lambda)m+\lambda = 0$$ To find the values of m, we will use the quadratic formula with \(a=1\), \(b=1+\lambda\), and \(c=\lambda\): $$m_{1,2} = \frac{-(1+\lambda)\pm\sqrt{(1+\lambda)^2-4\lambda}}{2}$$

Write the general solution

The general solution of the given differential equation is a linear combination of the solutions obtained from the values of m: $$y(x) = C_1e^{m_1x} + C_2e^{m_2x}$$ Now, we will apply the given boundary conditions to determine the eigenvalues and eigenfunctions.

Apply the first boundary condition

Apply the boundary condition \(y'(0)=0\): $$y'(x) = C_1m_1 e^{m_1x} + C_2m_2 e^{m_2x}$$ $$y'(0) = C_1m_1 + C_2m_2 = 0$$

Apply the second boundary condition

Apply the boundary condition \(y(1)=0\): $$y(1) = C_1e^{m_1} + C_2e^{m_2} = 0$$ Now we have a homogeneous system of linear equations: $$ \begin{cases} C_1m_1 + C_2m_2 &= 0 \\ C_1e^{m_1} + C_2e^{m_2} &= 0 \end{cases} $$ For a nontrivial solution, the determinant of the system must be zero: $$\begin{vmatrix} m_1 & m_2 \\ e^{m_1} & e^{m_2} \end{vmatrix} = 0$$

Determine the eigenvalues and eigenfunctions

Solve the homogeneous system of linear equations to find the eigenvalues and eigenfunctions. The eigenvalues \(\lambda\) can be obtained by solving the determinant equation of the linear system. Unfortunately, there is no straightforward closed-form expression for \(\lambda\) as a function of m. We must resort to numerical methods to find the eigenvalues. Once the eigenvalues are found, plug them back into the general solution to obtain the corresponding eigenfunctions. The eigenfunctions will be of the form: $$y_n(x) = C_1e^{m_1x} + C_2e^{m_2x}$$ where n denotes the index of the eigenvalue-eigenfunction pair.

Key concepts

Differential Equations

When studying differential equations, one dives into the realm of relationships between functions and their derivatives. In essence, a differential equation is a mathematical equation that links some function with its derivatives — a way of describing, in a compact form, how functions change.

Consider the physics of motion: It can be represented by differential equations that relate position, velocity (the first derivative of position with respect to time), and acceleration (the second derivative). In a similar sense, the textbook exercise in question presents a second-order differential equation where the function's derivatives up to the second order (notated as y'' and y') are involved.

Key Aspects:

• A second-order differential equation includes derivatives of the function up to the second degree.
• The notion of a parameter λ included in the differential equation adds an additional layer of complexity, often leading to a discussion of eigenvalues and eigenfunctions, especially when solving boundary value problems.
• Finding a solution typically involves hypothesizing a function form and deriving its implications.

Understanding how to approach and solve various differential equations is paramount for modeling many natural and engineered systems.

Boundary Value Problem

Boundary value problems (BVP) are a type of differential equation coupled with a set of additional constraints—boundary conditions. These problems require finding a function that satisfies not only the differential equation but also the conditions prescribed at the boundaries of the domain, which are often the edges of an interval.

Turning our attention to the exercise provided, the prescribed boundary conditions were y'(0) = 0 and y(1) = 0. Such conditions are crucial as they allow the analyst to narrow down the possible solutions to the differential equation, filtering out the functions that meet these end-point requisites.

Important elements to note:

• Initial conditions typically specify the value of the function and its derivatives at the beginning of the interval.
• Boundary conditions like those in our exercise can help pinpoint specific solutions, referred to as eigenfunctions, associated with their eigenvalues.
• Understanding and applying boundary conditions correctly is essential for solving physical problems, for example, determining the vibrational modes of a string fixed at both ends.

This conceptual framing is key to tackling a range of applied mathematics problems, especially in physics and engineering.

Linear Homogeneous Differential Equation

Diving into the concept of a linear homogeneous differential equation, we encounter a special class of equations. They are 'linear' because the function and its derivatives appear to the first power and are summed up linearly. The term 'homogeneous' implies that these equations equal zero when all terms are moved to one side, just like in the exercise's differential equation y'' + (1 + λ)y' + λy = 0.

A vital property of linear homogeneous equations is the principle of superposition, which lets us combine solutions linearly to generate new solutions. This agrees with the step from our exercise where a general solution is proposed as a linear combination of solutions for different values of m.

Essential points:

• A homogeneous differential equation equals zero when isolated on one side.
• Superposition allows for the linear combination of solutions.
• Finding solutions often starts with assuming a form, such as an exponential function, and working out implications to fit the equation and boundary conditions.

Understanding this category of differential equations broadens the approach to solving more complex problems, including those with boundary conditions, where eigenvalues and eigenfunctions play a significant role.