Question

Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+3 y=\cos x, \quad y^{\prime}(0)=0, \quad y^{\prime}(\pi)=0\)

Short answer

Based on the given second-order linear differential equation with boundary conditions, we have obtained the general solution: $$y(x) = A\cos{\sqrt{3}x} +\frac{1}{2}\cos x $$ However, as the second boundary condition does not uniquely determine the value of A, there are infinitely many solutions. The parameter A remains an arbitrary constant.

Step-by-step solution

Find the complementary function y_c for the homogeneous equation

The given differential equation is \(y^{\prime \prime} + 3y = \cos x\). First, we will find the complementary function y_c by solving the homogeneous equation \(y^{\prime \prime} + 3y = 0\). This leads to the characteristic equation \(r^2 + 3 = 0\). Solving, we find \(r=\pm \sqrt{-3}\). Therefore, the complementary function is given by: $$y_c(x) = A\cos{\sqrt{3}x} + B\sin{\sqrt{3}x}$$ where A and B are constants.

Find the particular solution y_p for the nonhomogeneous equation

Next, we will find a particular solution y_p for the nonhomogeneous equation. Since the nonhomogeneous term is \(\cos x\), we will assume a particular solution of the form: $$y_p(x) = C\cos x + D\sin x$$ Taking the first and second derivatives of y_p, we have: $$y_p^{\prime}(x) = -C\sin x + D\cos x$$ $$y_p^{\prime \prime}(x) = -C\cos x - D\sin x$$ Substituting y_p and its derivatives into the original differential equation, we get: $$(-C\cos x - D\sin x) + 3(C\cos x + D\sin x) = \cos x$$ This can be rewritten as: $$(-C+3C)\cos x + (-D+3D)\sin x = \cos x$$ Comparing coefficients, we get the following system of equations: $$2C = 1 \\ 2D = 0$$ Solving, we obtain C = 1/2 and D = 0. Therefore, the particular solution is: $$y_p(x) =\frac{1}{2}\cos x$$

Combine the complementary function and particular solution

Now, we combine the complementary function y_c and the particular solution y_p to form the general solution: $$y(x) = A\cos{\sqrt{3}x} + B\sin{\sqrt{3}x} +\frac{1}{2}\cos x $$

Apply the boundary conditions

We are given the boundary conditions \(y^{\prime}(0) = 0\) and \(y^{\prime}(\pi) = 0\). Taking the derivative of y(x): $$ y^{\prime}(x) = -A\sqrt{3}\sin{\sqrt{3}x} + B\sqrt{3}\cos{\sqrt{3}x} -\frac{1}{2}\sin x$$ Applying the first boundary condition \(y^{\prime}(0) = 0\): $$0 = -0 + B\sqrt{3} - 0$$ This gives us B=0. The second boundary condition is \(y^{\prime}(\pi) = 0\), which leads to: $$ 0 = A\sqrt{3}\sin(\sqrt{3}\pi) - \frac{1}{2}\sin(\pi)$$ Since \(\sin(\pi) = 0\), we are left with the equation: $$0 = A\sqrt{3}\sin(\sqrt{3}\pi)$$ We cannot determine A from this equation because \(\sin(\sqrt{3}\pi)\) does not equal 0, meaning this term may or may not cancel out depending on the value of A. Therefore, our final solution is: $$y(x) = A\cos{\sqrt{3}x} +\frac{1}{2}\cos x $$ The boundary value problem has infinitely many solutions, with A as an arbitrary constant.

Key concepts

Differential Equations

Understanding differential equations is fundamental to solving many problems in physics, engineering, and applied mathematics. These equations involve functions and their derivatives, providing a description of how a certain quantity changes over time or space. The equation essentially relates the rate of change of a function to the function itself.

For instance, the differential equation from the exercise, \(y^{\prime\prime} + 3y = \cos x\), is a second-order linear differential equation with non-constant coefficients. The order of the equation is determined by the highest derivative present, which in this case is the second derivative \(y^{\prime\prime}\). A linear differential equation is one where the function \(y\) and its derivatives appear to the power of one and are not multiplied together. The presence of the \(\cos x\) term makes this a nonhomogeneous differential equation, suggesting that the equation includes an external force or influence.

These equations are integral to modeling physical systems, and solving them provides insights into the behavior of the system. The general solution of a second-order linear differential equation such as this one consists of two parts: the complementary function \(y_c\), which is the solution of the homogeneous equation (when the equation equals zero), and the particular solution \(y_p\), corresponding to the nonhomogeneous part (the \(\cos x\) term in our example).

Complementary Function

The complementary function, \(y_c\), represents the solution to the homogeneous part of a differential equation, where the nonhomogeneous term (e.g., \(\cos x\)) is absent. In simpler terms, it is the general solution to the equation when set equal to zero, which describes the system's behavior without any external forces.

In our example, the homogeneous equation is \(y^{\prime\prime} + 3y = 0\). To find the complementary function, we solve the associated characteristic equation, which in this case leads to complex roots \(r = \pm\sqrt{-3}\). Therefore, the complementary function is a combination of sine and cosine functions: \[y_c(x) = A\cos{\sqrt{3}x} + B\sin{\sqrt{3}x}\]. Here, \(A\) and \(B\) are arbitrary constants that will later be determined by the boundary conditions provided.

The complementary function plays a critical role in the structure of the overall solution. It captures the natural dynamics of the system under study, which can be seen in phenomena like damped harmonic oscillators in mechanical systems or currents in electrical circuits.

Particular Solution

Moving beyond the effects encapsulated by the complementary function, we encounter the particular solution \(y_p\). This solution directly addresses the nonhomogeneous portion of the differential equation, representing the system's response to external forces or influences.

In the case of our exercise, the particular solution is investigated since the equation includes a nonhomogeneous term, \(\cos x\). We search for a \(y_p\) that satisfies the entire equation, not just the homogeneous part. The form of \(y_p\) is usually chosen based on the form of the nonhomogeneous term. Thus, for a \(\cos x\) term, a trial solution may include \(\cos x\) and \(\sin x\) components, resulting in \(y_p(x) = C\cos x + D\sin x\).

Solving for \(y_p\) involves substituting it into the original differential equation and then determining constants from the resulting equations. This process is known as the method of undetermined coefficients. For the given problem, the particular solution turns out to be \(\frac{1}{2}\cos x\), with \(C = \frac{1}{2}\) and \(D = 0\). The particular solution, in conjunction with the complementary function, forms the general solution to the original differential equation. When combined with the boundary conditions provided, these solutions can yield a specific solution for a particular problem or scenario.