Question

In Exercises 12-15 find a formula for the solution of the boundary problem by the method used to prove Theorem \(13.1 .3 .\) Assume that \(a

Short answer

In conclusion, the general solution for the given second-order linear ordinary differential equation (ODE) is: $$ y(x) = C_1e^x + C_2e^{-x} + e^x \int_{a}^{x} F(s)e^{s} ds + e^{-x} \int_{x}^{b} F(s)e^{-s} ds $$ The coefficients C1 and C2 can be determined by evaluating the integrals and finding the values of C1 and C2 that satisfy specific boundary conditions. The method used in this solution process consists of finding the complementary function, applying the method of variation of parameters (Undetermined Coefficients), and determining the coefficients for the boundary conditions.

Step-by-step solution

Find the Complementary Function

First, let's find the complementary function of the given ODE: $$ y^{\prime\prime} - y = 0 $$ This is a homogeneous linear ODE with constant coefficients. The auxiliary equation (characteristic equation) for this ODE is: $$ m^2 - 1 = 0 $$ Solving for m, we get \(m = \pm 1\). Thus, the general solution for the complementary function is: $$ y_c(x) = C_1e^x + C_2e^{-x} $$

Apply the Method of Variation of Parameters (Undetermined Coefficients)

Now, we'll apply the method of variation of parameters to find the particular solution. This method involves modifying the complementary function by introducing two new unknown functions, \(v_1(x)\) and \(v_2(x)\), as follows: $$ y_p(x) = v_1(x)e^x + v_2(x)e^{-x} $$ Next, we'll differentiate \(y_p(x)\) with respect to x once and twice. $$ y_p'(x) = v_1'(x)e^x + v_1(x)e^x - v_2'(x)e^{-x} + v_2(x)e^{-x} $$ $$ y_p''(x) = v_1''(x)e^x + 2v_1'(x)e^x + v_1(x)e^x + v_2''(x)e^{-x} - 2v_2'(x)e^{-x} + v_2(x)e^{-x} $$ Now, we substitute \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) into the given ODE: $$ y_p''(x) - y_p(x) = F(x) $$ $$ [v_1''(x)e^x + 2v_1'(x)e^x + v_1(x)e^x + v_2''(x)e^{-x} - 2v_2'(x)e^{-x} + v_2(x)e^{-x}] - [v_1(x)e^x + v_2(x)e^{-x}] = F(x) $$ Now, we need to satisfy the given ODE. To do this, we'll apply the following conditions: $$ v_1'(a) = 0 \quad and \quad v_2'(b)=0 $$ By substituting these conditions into the above equation, we'll have two separate equations: $$ v_1''(x)e^x + 2v_1'(x)e^x = \int_{a}^{x} F(s)e^{s} ds $$ $$ v_2''(x)e^{-x} - 2v_2'(x)e^{-x} = -\int_{x}^{b} F(s)e^{-s} ds $$ Now we can solve these two equations for \(v_1(x)\) and \(v_2(x)\). This will lead to the following particular solution: $$ y_p(x) = e^x \int_{a}^{x} F(s)e^{s} ds + e^{-x} \int_{x}^{b} F(s)e^{-s} ds $$ Thus, the general solution for the given ODE is: $$ y(x) = y_c(x) + y_p(x) = C_1e^x + C_2e^{-x} + e^x \int_{a}^{x} F(s)e^{s} ds + e^{-x} \int_{x}^{b} F(s)e^{-s} ds $$

Determine the Coefficients for the Boundary Conditions

Finally, to satisfy the boundary conditions \(y^{\prime}(a) = 0\) and \(y^{\prime}(b) = 0\), we will differentiate the general solution with respect to x and then substitute the boundary conditions. However, given that the boundary conditions depend on the function F(x) and we do not have a specific form for F(x), we cannot provide an explicit formula for the coefficients C1 and C2. Instead, we have the general solution for the given ODE: $$ y(x) = C_1e^x + C_2e^{-x} + e^x \int_{a}^{x} F(s)e^{s} ds + e^{-x} \int_{x}^{b} F(s)e^{-s} ds $$ Any particular solution with specific boundary conditions can be found by evaluating the integrals and finding the values of C1 and C2 that satisfy the boundary conditions.

Key concepts

Complementary Function

In the context of solving differential equations, the complementary function, denoted as y_c(x), represents the solution of the corresponding homogeneous differential equation. To elaborate, when we have an equation like

\[y'' - y = F(x),\]
the complementary function is derived by setting the right side to zero:

\[y'' - y = 0.\]
The solution to this homogeneous equation provides the basis of the general solution to the original differential equation with the non-homogeneous term, F(x).The process involves first determining the characteristic equation, which in this case is a quadratic equation formed by assuming the solution is of the form e^mx. Here, it is

\[m^2 - 1 = 0.\]
Solving it yields the roots m = \text{\textpm} 1, leading to the complementary function:

\[y_c(x) = C_1e^x + C_2e^{-x}.\]
This part of the solution handles the behavior of the equation without the influence of the forcing function, F(x), and sets the stage for finding the particular solution, which includes the effect of F(x).

Method of Variation of Parameters

One technique to solve non-homogeneous differential equations like

\[y'' - y = F(x)\]
is the Method of Variation of Parameters. After foundational work that includes finding the complementary function, variation of parameters is a process wherein we seek a particular solution y_p(x) by varying the parameters of the complementary function. The method is especially useful when the non-homogeneous term does not easily lend itself to simpler methods such as undetermined coefficients.We introduce two functions, v₁(x) and v₂(x), which will replace the constants C₁ and C₂, assuming a form for y_p analogous to y_c:

\[y_p(x) = v_1(x)e^x + v_2(x)e^{-x}.\]
Differentiating and substituting back into the original equation allows us to establish a system of equations involving v₁' and v₂'. From there, we can often integrate to find expressions for these functions, leading to the particular solution that accounts for the influence of the non-homogeneous term F(x).

Characteristic Equation

A crucial step when dealing with linear homogeneous differential equations is forming and solving the characteristic equation. For our example

\[y'' - y = 0,\]
we assume that the solution to the differential equation can be expressed as e^mx. We substitute this into the homogeneous part of the equation, leading to the characteristic equation

\[m^2 - 1 = 0.\]
Typically, a quadratic equation in m produced from a second-order linear homogeneous differential equation will give us the roots that are necessary to construct the general solution. In the characteristic equation, m₁ and m₂ (which are in our case) indicate the nature of the solution—real and distinct, real and repeated, or complex conjugates. This dictates whether our complementary function will consist of exponential functions, sine and cosine functions, or a combination of both.

Homogeneous Linear ODE

A homogeneous linear ordinary differential equation (ODE) is one in which the function y(x) and its derivatives appear linearly and there are no terms that are functions of x alone. Essentially, an n^th-order homogeneous linear ODE can be written in the general form

\[a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_1(x)y' + a_0(x)y = 0.\]
In our specific case, the equation

\[y'' - y = 0\]
represents a second-order homogeneous linear ODE with constant coefficients. These equations are particularly approachable due to their constant coefficient characteristic, which affords us the luxury of predicting their solutions' form. The concept of superposition applies here, meaning that the sum of any two solutions to the homogeneous equation is also a solution—a principle that enormously simplifies the task of finding the general solution.