Question

Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+2 y=x, \quad y(0)=0, \quad y(\pi)=0\)

Short answer

Question: Determine whether the boundary value problem \(y'' + 2y = x\), \(y(0) = 0\), and \(y(\pi) = 0\) has a unique solution and, if so, find the solution. Solution: The boundary value problem has a unique solution, given by: \[y(t) = -\frac{1}{2}\pi\frac{\sin(\sqrt{2} t)}{\sin(\sqrt{2}\pi)} + \frac{1}{2}t\]

Step-by-step solution

Rewrite the ODE in standard form

First, let's rewrite the given ODE in standard form: \[y'' + 2y = x\]

Find the complementary function

We can find the complementary function by solving the homogeneous ODE associated with the given ODE: \[y'' + 2y = 0\] Since this is a linear homogeneous ODE with constant coefficients, we can propose a solution in the form of \(y_c = e^{rt}\), where \(r\) is a constant. Substituting this into the homogeneous ODE, we get \[r^2 e^{rt} + 2e^{rt} = 0\] Factoring out the common factor of \(e^{rt}\), we have \[(r^2 + 2) e^{rt} = 0\] Since \(e^{rt}\) is never zero, the term in parentheses must be zero: \[r^2 + 2 = 0\] This equation has two complex roots: \(r_1 = i\sqrt{2}\) and \(r_2 = -i\sqrt{2}\). Thus, the complementary function is given by \[y_c(t) = A\cos(\sqrt{2} t) + B\sin(\sqrt{2} t)\]

Find the particular solution

Now, we need to find a particular solution to the given ODE. Since the right-hand side of the equation is a polynomial function, we can propose the particular solution \(y_p(t) = At + B\). Differentiating twice with respect to \(t\), we get \[y'_p(t) = A\] \[y''_p(t) = 0\] Substituting the particular solution and its derivatives into the given ODE, we have \[0 + 2(At + B) = x\] Comparing coefficients, we find that \(A = \frac{1}{2}\) and \(B = 0\). Thus, our particular solution is \[y_p(t) = \frac{1}{2}t\]

Form the general solution

Our general solution is the sum of the complementary function and the particular solution: \[y(t) = y_c(t) + y_p(t)\] \[y(t) = A\cos(\sqrt{2} t) + B\sin(\sqrt{2} t) + \frac{1}{2}t\]

Apply the boundary conditions

Now, apply the boundary conditions to the general solution: \[y(0) = A\cos(0) + B\sin(0) + \frac{1}{2}(0) = 0\] \[A = 0\] Next, apply the second boundary condition: \[y(\pi) = 0\cos(\sqrt{2}\pi) + B\sin(\sqrt{2}\pi) + \frac{1}{2}\pi = 0\] Solving for B, we get \[B\sin(\sqrt{2}\pi) = -\frac{1}{2}\pi\] Since sin is not zero for \(\sqrt{2}\pi\), the boundary value problem has a unique solution, which is: \[y(t) = -\frac{1}{2}\pi\frac{\sin(\sqrt{2} t)}{\sin(\sqrt{2}\pi)} + \frac{1}{2}t\]

Key concepts

Boundary Value Problem

In mathematics, a Boundary Value Problem (BVP) involves finding a solution to a differential equation that not only satisfies the equation itself but also the boundary conditions. Imagine it like setting up a story; you not only have the primary plot (differential equation) but also specific conditions at the start and end (boundary values).

For our problem, we see:

• The differential equation: \(y'' + 2y = x\)
• Boundary conditions: \(y(0) = 0\) and \(y(\pi) = 0\)

The challenge is to find a function \(y\) that adheres to both the equation and these boundary constraints. It’s crucial because the boundary conditions ensure our solution is tailor-made for specific scenarios. Boundary value problems often arise in physics and engineering since real-world problems typically come with constraints or conditions to meet.

Complementary Function

The Complementary Function (CF) plays a major role in solving linear differential equations. It originates from the homogeneous part of the differential equation, meaning the piece of the equation without the external input or forcing function.

Consider: \(y'' + 2y = 0\), which is the homogeneous version of our original equation. Solving this gives the complementary function, which covers the natural response of the system to initial conditions.

• Assume a solution \(y_c = e^{rt}\)
• Equations like this often have solutions in the form of exponentials.
• Finding \(r\), in our case, results in complex roots, due to the equation \(r^2 + 2 = 0\).

For the given complex roots \(r_1 = i\sqrt{2}\) and \(r_2 = -i\sqrt{2}\), the CF becomes: \(y_c(t) = A\cos(\sqrt{2} t) + B\sin(\sqrt{2} t)\). This accounts for the part of the solution characterized by oscillatory (sine and cosine) functions, reflecting the continuous and periodic nature of physical systems.

Particular Solution

The Particular Solution (PS) differs because it specifically tackles the inhomogeneous part of the differential equation, essentially taking care of any external input or driving forces present on the right-hand side of the equation.

In our equation \(y'' + 2y = x\), the right side \(x\) acts as an external force. To find the particular solution:

• Assume a form based on the external input: For polynomial \(x\), assume \(y_p(t) = At + B\)
• Differentiate and substitute back into the original equation to determine coefficients.

For our problem, solving gives \(A = \frac{1}{2}\) and \(B = 0\), so the particular solution is \(y_p(t) = \frac{1}{2}t\). This part of the solution models the specific impact of the external force or condition on the system.

Complex Roots

When dealing with differential equations, encountering complex roots is quite common, especially in constant coefficient linear homogeneous equations. Complex roots tend to imply oscillatory behavior within the solution, usually involving trigonometric expressions.

For the problem \(y'' + 2y = 0\), solving for characteristic equation \(r^2 + 2 = 0\) yields complex roots \(r_1 = i\sqrt{2}\) and \(r_2 = -i\sqrt{2}\). These complex roots lead us to a solution involving sine and cosine:

• The imaginary part of the root translates to the frequency of the oscillation (\(\sqrt{2}\) in our case).
• This results in the complementary function: \(y_c(t) = A\cos(\sqrt{2} t) + B\sin(\sqrt{2} t)\).

This type of solution is essential because it reflects systems that naturally oscillate, which is pivotal in fields like acoustics, electrical circuits, and mechanical systems.