Question

Either solve the given boundary value problem or else show that it has no solution. \(y^{\prime \prime}+y=0, \quad y(0)=0, \quad y^{\prime}(\pi)=1\)

Short answer

Given a boundary value problem involving the second-order linear homogeneous differential equation \(y''+y=0\) with boundary conditions \(y(0)=0\) and \(y'(\pi)=1\), it was found that the general solution is given by \(y(x) = A \cos(x) + B \sin(x)\). However, applying the boundary conditions resulted in an inconsistent solution, ultimately leading to the conclusion that the problem has no solution due to the inconsistency of the given boundary conditions with the given differential equation.

Step-by-step solution

Check the consistency of boundary conditions

First, we need to check if the boundary conditions are consistent with the given differential equation. If they are, we can proceed to find the solution of the problem. The given boundary conditions are: \(y(0)=0\) and \(y'(\pi)=1\) The given differential equation is: \(y'' + y = 0\) Notice that the differential equation is linear and homogeneous, meaning that if \(\phi\) and \(\psi\) are solutions, then any linear combination of those solutions is also a solution. However, there's no way we can find a linear combination such that the derivatives at \(\pi\) would equal \(1\). Hence, the boundary conditions are inconsistent, and the problem has no solution. Alternatively, in the next steps, we'll try to find the general solution and use the boundary conditions to see if we can find a particular solution.

Find the general solution

The given differential equation is: \(y''+y=0\) This is a second-order linear homogeneous differential equation with constant coefficients. We can first try a solution of the form \(y=e^{rx}\). Plugging this into the differential equation, we get: \((re^{rx})^2 + e^{rx} = 0\) Divide both sides by the non-zero term \(e^{rx}\): \(r^2 + 1 = 0\) This is a quadratic equation in \(r\). Solving for \(r\), we get: \(r^2 = -1\) \(r=\pm i\) Now, we can write the general solution in terms of sines and cosines using Euler's formula: \(y(x) = A \cos(x) + B \sin(x)\)

Apply boundary conditions

Now, we'll try to apply the boundary conditions and see if we can find a particular solution. Boundary condition 1: \(y(0) = 0\) We get: \(0 = A \cos(0) + B \sin(0)\) \(0 = A\) Boundary condition 2: \(y'(\pi) = 1\) We first need to find the expression for the derivative of \(y'(x)\): \(y'(x) = -A\sin(x) + B \cos(x)\) Now, plug in the boundary condition \(y'(\pi) = 1\): \(1 = -A \sin(\pi) + B \cos(\pi)\) Given that \(A = 0\), we have: \(1 = -B\) Now we have \(A=0\) and \(B=-1\), so the particular solution of the differential equation would be \(y(x) = -\sin(x)\). However, this particular solution does not satisfy the second boundary condition: \(y'(\pi)=1\). Instead, we find that the derivative at this value is \(y'(\pi)=-1\), not \(1\). Hence, we can conclude that the problem has no solution since the given boundary conditions are inconsistent with the given differential equation.

Key concepts

Linear Homogeneous Differential Equation

Linear homogeneous differential equations are equations where the function and its derivatives appear linearly, and all terms involve the unknown function. An example of this is \( y'' + y = 0 \), our given equation. Here, "homogeneous" indicates that the equation equals zero, meaning no extra terms are present. The solutions to these equations often involve finding specific functions that satisfy the equation when plugged back in.

• These equations have the form \( a_n\frac{d^n y}{dx^n} + a_{n-1}\frac{d^{n-1} y}{dx^{n-1}} + \ldots + a_1\frac{dy}{dx} + a_0 y = 0 \).
• The coefficients \(a_n, a_{n-1}, \ldots, a_0\) are typically constants.
• Thus, they are ideal for systems with consistent rates of change, modeling phenomena like oscillations.

Solving these equations often comes down to finding exponential, trigonometric, or polynomial functions that fit the criteria set by the equation structure.

Boundary Conditions

Boundary conditions define the specifics of a solution at particular values of the independent variable, usually at the endpoints of an interval. In our case, we have:

• \( y(0) = 0 \), which tells us the solution must pass through the origin at \( x = 0 \).
• \( y'(\pi) = 1 \), which specifies the slope of the solution at \( x = \pi \).

In boundary value problems, these conditions are crucial as they guide the specificity of the solution beyond just the general form.
Sometimes, the boundary conditions can reveal inconsistencies, as in our example. The conditions can be such that no single function satisfies both, indicating no solution exists. Investigating these limitations helps us understand the conditions under which a physical or theoretical system aligns with reality or needs reevaluation.

Trigonometric Solutions

Trigonometric solutions often appear in differential equations where the characteristic equation yields complex roots. For the equation \( y'' + y = 0 \), we derived the general solution to be \( y(x) = A \cos(x) + B \sin(x) \). These solutions arise due to the periodic nature of trigonometric functions:

• \( \cos(x) \) and \( \sin(x) \) are fundamental solutions often used when equations involve simple harmonic motion.
• They provide physically meaningful representations for oscillatory systems, such as springs or circuits.

In this particular boundary value problem, these trigonometric forms align well with the differential equation but fail to satisfy the inconsistent boundary conditions. Therefore, while powerful for periodic functions, trigonometric solutions require consistent conditions to be valid.

Euler's Formula

Euler's Formula is a pivotal equation in mathematics that bridges exponential and trigonometric functions. It states that \( e^{ix} = \cos(x) + i \sin(x) \) for any real number \( x \). This formula is essential in solving differential equations with complex roots, as it allows us to express complex exponentials as real sine and cosine functions.

• Using Euler's formula, the roots \( r = \pm i \) transform into solutions involving \( \cos(x) \) and \( \sin(x) \).
• It elegantly shows how rotations in the complex plane correspond to trigonometric oscillations.

In our example, applying Euler's formula helped rewrite the general solution in a real and usable form: \( y(x) = A \cos(x) + B \sin(x) \). This form makes it easier to apply and check against given boundary conditions. However, even with these powerful tools, the given scenario's boundary conditions did not allow for a consistent solution.