Question

In Exercises \(2-7\) solve the boundary value problem. $$ y^{\prime \prime}+4 y=1, \quad y(0)=3, \quad y(\pi / 2)+y^{\prime}(\pi / 2)=-7 $$

Short answer

Question: Determine the specific solution for the boundary value problem \(y'' + 4y = 1\), with boundary conditions \(y(0) = 3\) and \(y(\pi / 2) + y'(\pi / 2) = -7\). Answer: The specific solution for the given boundary value problem is \(y(x) = \frac{11}{4}\cos(2x) + \frac{1}{2}\sin(2x) + \frac{1}{4}\).

Step-by-step solution

1. Complementary Function

To find the complementary function, we need to solve the homogeneous version of the given ODE: \(y'' + 4y = 0\). This is a linear ODE with constant coefficients. We can set up the characteristic equation: \(r^2 + 4 = 0\). Solving for \(r\), we get complex roots \(r = \pm 2i\). The general form of the complementary function is given by: \(y_c(x) = A\cos(2x) + B\sin(2x)\), where \(A\) and \(B\) are constants.

2. Particular Integral

We now find a particular integral for the given ODE. Since the right side of the ODE is a constant, we'll assume the particular integral is also a constant, \(y_p(x) = C\). Taking the first and second derivatives: \(y_p'(x) = 0\) and \(y_p''(x) = 0\). Plugging this back into the original ODE: \(0 + 4C = 1 \Rightarrow C = 1/4\). So the particular integral is \(y_p(x) = \frac{1}{4}\).

3. General Solution

The general solution of the given ODE combines the complementary function (\(y_c(x)\)) and the particular integral (\(y_p(x)\)): \(y(x) = y_c(x) + y_p(x) = A\cos(2x) + B\sin(2x) + \frac{1}{4}\).

4. Applying the First Boundary Condition

Now we apply the given boundary conditions. The first one is \(y(0) = 3\). Plugging in \(x = 0\) into the general solution, we get: \(3 = A\cos(0) + B\sin(0) + \frac{1}{4} \Rightarrow 3 = A + 0 + \frac{1}{4}\). Solving for \(A\), we find \(A = \frac{11}{4}\).

5. Applying the Second Boundary Condition

Now, we apply the second boundary condition, \(y(\pi / 2) + y'(\pi / 2) = -7\). First, we need to find the derivative of the general solution: \(y'(x) = -2A\sin(2x) + 2B\cos(2x)\). Now, we plug in \(x = \pi/2\) for \(y(x)\) and \(y'(x)\): \(y(\pi / 2) = \frac{11}{4}\cos(\pi) + B\sin(\pi) + \frac{1}{4} = -\frac{11}{4} + \frac{1}{4}\). \(y'(\pi / 2) = -2(\frac{11}{4})\sin(\pi) + 2B\cos (\pi) = -2B\). Applying the boundary condition, we get: \( -\frac{10}{4} - 2B = -7 \Rightarrow B = \frac{1}{2} \).

6. Specific Solution

We now have the values of \(A\) and \(B\). Plugging them back into the general solution, we get the specific solution for the given BVP: \(y(x) = \frac{11}{4}\cos(2x) + \frac{1}{2}\sin(2x) + \frac{1}{4}\).

Key concepts

Complementary Function

The complementary function is crucial in solving a boundary value problem, especially when dealing with a linear ordinary differential equation (ODE) with constant coefficients. It is essentially the solution to the homogeneous equation, meaning where the equation equals zero. When we are given an ODE like
\[ y''+4y=0, \]
we find the solution that corresponds to the set of conditions when the equation is not influenced by any external forces or inputs—in other words, the natural response of the system.

To find this solution, we use characteristic equations derived from the ODE. In our example, the characteristic equation \[ r^2 + 4 = 0 \] leads to complex roots. The solutions to such equations with complex roots are sinusoidal functions. This is why our complementary function takes the form \[ y_c(x) = A\cos(2x) + B\sin(2x), \] where \(A\) and \(B\) are constants that will be determined by the boundary conditions of the problem.

Particular Integral

While the complementary function represents the system's homogeneous response, the particular integral is the solution that specifically addresses the non-homogeneous part of the ODE—in our case, the constant on the right-hand side of \[ y''+4y=1. \] To find the particular integral, we make an educated guess based on the form of the non-homogeneous term. Because our non-homogeneous term is a constant, we assume a constant solution for the particular integral, denoted as \(y_p(x) = C\).

When this constant particular integral is substituted into the ODE, we solve for \(C\) to balance the equation. With this method, we determined that \[ y_p(x) = \frac{1}{4}. \] This particular integral will be used, along with the complementary function, to construct the general solution to the original ODE.

Linear ODE with Constant Coefficients

The boundary value problem given involves a linear ODE with constant coefficients, which means that each term is a derivative of \(y\) (or \(y\) itself), multiplied by a constant, and summed to equal a function of \(x\). Such equations are especially manageable because their general solutions combine the complementary function (from the associated homogeneous equation) and the particular integral (from the non-homogeneous equation).

The distinct feature of constant coefficients allows us to use specific methods, like characteristic equations, to find solutions quickly. This property also guarantees the existence of a set of basic solutions that can be linearly combined to create the general solution. We use boundary conditions to define the particular coefficients for our specific case, arriving at the specific solution to fully describe the behavior delineated by the boundary value problem.

Following these steps is not only critical for solving the equation but also builds intuition for understanding the underlying physical or abstract processes that are described by such linear ODEs with constant coefficients.