Question

Determine whether there is any value of the constant \(a\) for which the problem has a solution. Find the solution for each such value. $$ y^{\prime \prime}+\pi^{2} y=a, \quad y^{\prime}(0)=0, \quad y^{\prime}(1)=0 $$

Short answer

If yes, provide the solution in terms of \(a\). Given ODE and boundary conditions: $$ y^{\prime \prime} + \pi^{2} y = a, \quad y^{\prime}(0) = 0, \quad y^{\prime}(1) = 0 $$ Solution: The problem has a solution for any value of \(a\), with the solution given as: $$ y(x) = \frac{a}{\pi^2}. $$

Step-by-step solution

Write down the given ODE and boundary conditions

The given differential equation with boundary conditions is: $$ y^{\prime \prime} + \pi^{2} y = a, \quad y^{\prime}(0) = 0, \quad y^{\prime}(1) = 0 $$

Solve the homogeneous ODE

The homogeneous version of the given ODE is: $$ y^{\prime \prime} + \pi^2 y = 0 $$ The characteristic equation is: $$ r^2 + \pi^2 = 0 \Rightarrow r = \pm \pi i, $$ Thus, the general solution of the homogeneous ODE is $$ y_h(x) = C_1 \cos{\pi x} + C_2 \sin{\pi x}. $$

Solve the particular solution

In order to find a particular solution, we can use the method of undetermined coefficients. Since the non-homogeneous term is a constant \(a\), we can assume that the particular solution is also a constant. Let \(y_p(x) = A\). Then, $$ y_p^{\prime}(x) = 0, \quad y_p^{\prime \prime}(x) = 0 $$ Substituting into the given ODE, we have: $$ 0 + \pi^2 A = a \Rightarrow A = \frac{a}{\pi^2} $$ Hence, the particular solution is: $$ y_p(x) = \frac{a}{\pi^2}. $$

Combine the homogeneous and particular solutions

The general solution of the given ODE is the sum of the homogeneous and particular solutions: $$ y(x) = y_h(x) + y_p(x) = C_1 \cos{\pi x} + C_2 \sin{\pi x} + \frac{a}{\pi^2}. $$

Apply boundary conditions and determine the possible values of \(a\)

We apply the boundary conditions \(y^{\prime}(0) = 0\) and \(y^{\prime}(1) = 0\). First, we compute the derivative of the general solution: $$ y^{\prime}(x) = -C_1 \pi \sin{\pi x} + C_2 \pi \cos{\pi x}. $$ Now we apply the boundary conditions: $$ y^{\prime}(0) = -C_1 \pi \sin{\pi(0)} + C_2 \pi \cos{\pi(0)} = C_2 \pi = 0 \Rightarrow C_2 = 0 $$ and $$ y^{\prime}(1) = -C_1 \pi \sin{\pi(1)} + 0 = -C_1 \pi = 0 \Rightarrow C_1 = 0 $$ So both \(C_1\) and \(C_2\) are equal to zero. The general solution now becomes $$ y(x) = \frac{a}{\pi^2}. $$ Since there are no conditions on \(a\), the problem has a solution for any value of \(a\). The solution for each value of \(a\) is: $$ y(x) = \frac{a}{\pi^2}. $$

Key concepts

Homogeneous ODE

Understanding the concept of a homogeneous ordinary differential equation (ODE) is central to solving many problems in differential equations. In essence, a homogeneous ODE is an equation that represents the relationship between a function and its derivatives, without any external forces or inputs—it has a zero on the right side of the equation.

When faced with a differential equation such as
$$y^{\' \'} + \time^2 y = 0,$$
the task is to find the function y(x) that satisfies this relationship for all x. Finding the solution involves searching for functions whose second derivative—when added to the product of the function itself and \time^2—equals zero.

The key feature of the solution to the homogeneous ODE is that it will involve arbitrary constants, since there are infinitely many functions that can satisfy the equation. Reflecting on the exercise, the general solution
$$y_h(x) = C_1 \cos{\pi x} + C_2 \sin{\pi x}$$
comprises two parts: one involving a cosine term and one with a sine term. This generic form will work for all values of C_1 and C_2, which ultimately are determined by the initial or boundary conditions supplied with the problem.

Particular Solution

The quest for a particular solution is an expedition into the realm of specificity. While the homogeneous solution deals with the general, the particular solution seeks to address the non-zero right-hand side of a differential equation. In the lay of this educational land, we aim to find a function that, when plugged into our non-homogeneous equation, perfectly balances the scales.

In our textbook example above, we are given
$$y^{\' \'} + \pi^2 y = a,$$and it's our mission to find this equilibrium-seeking 'particular' value. We propose a potential solution—let's call it y_p(x)—and determine its derivatives. Since our non-homogeneous term is a constant (in this case, a), we could hypothesize that y_p(x) may be constant as well.

Applying this particular solution to the differential equation simplifies our equation to
$$\pi^2 A = a$$and, voilà, we deduce that
$$y_p(x) = \frac{a}{\pi^2}.$$

This single value of A is a revelation, filling in the piece of the puzzle that the homogeneous solution left unknown and rendering a full picture of the differential equation's solutions.

Method of Undetermined Coefficients

The method of undetermined coefficients is akin to playing detective within the world of differential equations. It's a strategy we employ when seeking a particular solution to a non-homogeneous ordinary differential equation. This method is best used when the non-homogeneous term—typically on the right side of the equation—is relatively simple, like a polynomial, exponential, sine, or cosine function.

In our case, with the right side being the constant ‘a’, we hypothesized that our particular solution y_p(x) might also take the form of a constant. We name this constant ‘A’ and deem it ‘undetermined’ because we've yet to ascertain its value.

Then we calculate the derivatives of our proposed y_p(x), which are zero due to its constant nature. By substituting it into the differential equation, we are able to isolate and determine an explicit value for 'A', satisfying the non-homogeneous part of the equation. This method of undetermined coefficients simplifies the process of finding particular solutions, especially when the form of these solutions can be reasonably guessed at the outset.

Characteristic Equation

Peering into a crystal ball, the characteristic equation offers a glimpse of the potential solutions for the roots of a homogeneous ODE. It translates the differential equation into an algebraic one, which is much more docile and tractable. Essentially, defining the characteristic equation is a transformative act, converting the differential problem into one of solving for the roots of a polynomial, which are the characteristic values.

For the homogeneous equation
$$y^{\' \'} + \pi^2 y = 0,$$we write its characteristic equation as
$$r^2 + \pi^2 = 0.$$Then, after some algebraic manipulation, we discover the roots to be complex numbers $$r = \pm \pi i.$$

These roots tell us about the nature of the solutions to our ODE. Since they are purely imaginary, we expect the solutions to involve sine and cosine functions, which are moulded in the form of harmonic oscillations with constants that will be determined by the boundaries of our problem. Embarking on solving the characteristic equation provides us with the foundation of the general solution to the homogeneous ODE and paves the way to the complete solution, once combined with our particular solution.