Question

Solve the given problem by means of an eigenfunction expansion. $$ y^{\prime \prime}+2 y=-x, \quad y(0)=0, \quad y^{\prime}(1)=0 ; \quad \text { see Section } 11.2, \text { Problem } 7 $$

Short answer

Answer: The solution to the given boundary value problem is \(y(x) = -\frac{1}{2}x + \frac{1}{4}\).

Step-by-step solution

Find the homogeneous solution

First, we need to find the homogeneous solution of the given differential equation. To do this, we need to rewrite the equation in its homogeneous form, i.e., we set the right-hand side equal to zero: $$ y^{\prime \prime}_h + 2y_h = 0. $$ Now, we can easily solve for the homogeneous solution \(y_h(x)\). This is a simple second-order linear homogeneous differential equation with constant coefficients. The general solution is: $$ y_h(x) = C_1 e^{\sqrt{-2}x} + C_2 e^{-\sqrt{-2}x}. $$

Apply the boundary conditions to the homogeneous solution

Next, we need to apply the boundary conditions to the homogeneous solution \(y_h(x)\): 1. Apply the left boundary condition, \(y(0) = 0\): $$ 0 = C_1 e^{0} + C_2 e^{0} \implies C_1 + C_2 = 0. $$ 2. Apply the right boundary condition, \(y^{\prime}(1) = 0\): First, find the derivative: $$ y^{\prime}_h(x) = \sqrt{-2} C_1 e^{\sqrt{-2}x} - \sqrt{-2} C_2 e^{-\sqrt{-2}x}, $$ then apply \(y^{\prime}(1) = 0\): $$ 0 = \sqrt{-2} C_1 e^{\sqrt{-2}} - \sqrt{-2} C_2 e^{-\sqrt{-2}}. $$ Solving the system of two equations for \(C_1\) and \(C_2\), we get \(C_1 = C_2 = 0\), therefore: $$ y_h(x) = 0. $$

Find the particular solution

Now, we need to find the particular solution of the given nonhomogeneous differential equation. We can use the method of undetermined coefficients to guess a solution: $$ y_p(x) = ax + b. $$ Calculating its second derivative and substituting it into the equation, we get: $$ (2a) + 2(ax + b) = -x. $$ Comparing the coefficients of \(x\) and constants, we get the following two equations: $$ 2a = -1, \quad 2b + 2a = 0. $$ Solving the system of equations yields \(a = -\frac{1}{2}\) and \(b = \frac{1}{4}\). Therefore, the particular solution is: $$ y_p(x) = -\frac{1}{2}x + \frac{1}{4}. $$

Combine the homogeneous and particular solutions

Now, we combine the homogeneous and particular solutions to obtain the general solution of the given differential equation: $$ y(x) = y_h(x) + y_p(x) = 0 - \frac{1}{2}x + \frac{1}{4} = -\frac{1}{2}x + \frac{1}{4}. $$ Thus, the solution to the given boundary value problem using eigenfunction expansion is: $$ y(x) = -\frac{1}{2}x + \frac{1}{4}. $$

Key concepts

Homogeneous Differential Equation

When we talk about a homogeneous differential equation, we are referring to an equation that is set equal to zero throughout. For example, in the case of a second-order linear homogeneous differential equation, the general form is expressed as \( y'' + p(x)y' + q(x)y = 0 \), where \( p(x) \) and \( q(x) \) are functions of \( x \).

In our exercise, the homogeneous equation is \( y''_h + 2y_h = 0 \). Solving this kind of equation usually involves finding a characteristic equation based on the coefficients of \( y \) and its derivatives; in this case, the characteristic equation is a quadratic equation from which we determine the nature of the solutions based on its roots. Since the characteristic roots in this case are complex, the general solution involves exponential functions of \( x \) multiplied by the coefficients \( C_1 \) and \( C_2 \).

Understanding how to solve these equations is vital because the approach sets the foundation for finding the complete solution to more complex, nonhomogeneous equations, like the one in our problem. Moreover, in physical problems, the homogeneous solution often represents the natural behavior of the system without external forces acting upon it.

Boundary Value Problem

A boundary value problem is a type of differential equation along with a set of additional constraints called boundary conditions. The solution to the differential equation must satisfy these conditions at the boundaries of the domain of the problem, which is typically a range of values for \( x \).

In our exercise, the boundary conditions provided are \( y(0) = 0 \) and \( y'(1) = 0 \). These conditions are crucial for determining the specific coefficients \( C_1 \) and \( C_2 \) in the solution to the homogeneous equation. After applying these conditions, we find that the constants must both be zero, leading to the homogeneous solution \( y_h(x) = 0 \).

It's important to notice that boundary value problems are distinct from initial value problems, which involve conditions specified at a single point rather than across a range of the domain. Boundary value problems are particularly relevant in fields such as physics and engineering where conditions at the edges of a spatial domain (like the ends of a string or the surface of a drum) are known.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find a particular solution to a nonhomogeneous differential equation when the nonhomogeneous term is a linear combination of certain types of functions, such as polynomials, exponentials, or sine and cosine functions.

In the case of our problem, the nonhomogeneous part of the differential equation is a linear function of \( x \), so we guess a solution in the form \( y_p(x) = ax + b \). We then differentiate this guess, substitute back into the nonhomogeneous equation, and compare coefficients to find the values of \( a \) and \( b \).

This method is powerful because it provides a systematic way to tackle differential equations without delving into the complexity of integration, which is required in other methods like variation of parameters. It's essential, however, to correctly identify the form of the particular solution based on the nonhomogeneous term to ensure that this technique can be applied successfully.