Question

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

Short answer

Answer: Yes, there exists a solution. The particular solution is: $y(x) = \begin{cases} \cos(x) + \sin(x), & \textrm{if } L \textrm{ is not an integer multiple of } \pi \\ (-1)^n + \sin(x), & \textrm{if } L = n\pi, n \textrm{ is even} \\ (-1)^n - 1 + \sin(x), & \textrm{if } L = n\pi, n \textrm{ is odd} \end{cases}$

Step-by-step solution

Write down the given differential equation and boundary conditions

We have the differential equation: \(y''+y=0 \) And the boundary conditions: \(y'(0)=1\) \(y(L)=0\)

Solve the homogeneous differential equation

To solve the homogeneous differential equation, we should find the characteristic equation: \(r^2 + 1 = 0\) Solving for r, we get: \(r=\pm i\) The general solution for the homogeneous differential equation is then: \(y(x) = c_1 \cos(x) + c_2 \sin(x)\)

Apply boundary condition \(y'(0)=1\)

Differentiate the general solution with respect to x to find y'(x): \(y'(x) = -c_1 \sin(x) + c_2 \cos(x)\) Now apply the boundary condition \(y'(0)=1\): \(y'(0) = -c_1 \sin(0) + c_2 \cos(0) = c_2\) Thus, \(c_2 = 1\). So our equation becomes: \(y(x) = c_1 \cos(x) + \sin(x)\)

Apply boundary condition \(y(L)=0\)

Apply the second boundary condition \(y(L)=0\): \(y(L) = c_1 \cos(L) + \sin(L) = 0\) Now, we have to analyze if there is any value for \(c_1\) to satisfy the above equation. If \(L\) is not an integer multiple of \(\pi\), there will always be a unique \(c_1\) that satisfies the above equation, since \(\cos(L) \neq 0\). So a solution must exist. If \(L = n\pi\) where \(n\) is an integer: \(0 = c_1 \cos(n\pi) + \sin(n\pi) = c_1 (-1)^n\) In this case, the boundary condition \(y(L)=0\) reduces to: \(c_1 = 0\) if \(n\) is even, \(c_1 = -2\) if \(n\) is odd. Therefore, regardless of the value of \(L\), there exists a solution. In particular: $y(x) = \begin{cases} \cos(x) + \sin(x), & \textrm{if } L \textrm{ is not an integer multiple of } \pi \\ (-1)^n + \sin(x), & \textrm{if } L = n\pi, n \textrm{ is even} \\ (-1)^n - 1 + \sin(x), & \textrm{if } L = n\pi, n \textrm{ is odd} \end{cases}$

Key concepts

Differential Equations

Differential equations are mathematical tools we use to describe phenomena where change is ongoing, like population growth or the motion of objects. The equation given in the exercise, \(y^{\backprime \backprime}+y=0\), is a simple form of a differential equation where the change in \(y\) depends on \(y\) itself. To solve such an equation, we first identify the type of differential equation we're dealing with, which in this case is a second-order linear homogeneous differential equation with constant coefficients.

For students wanting to understand this better, think of a differential equation as a recipe that describes how one quantity changes in response to another. The solution to a differential equation is a function or a set of functions that satisfy the equation. In the context of the given problem, we are looking for a function \(y(x)\) that, when plugged into the differential equation, transforms it into a true statement.

Characteristic Equation

The characteristic equation is a tool we use to solve linear homogeneous differential equations with constant coefficients. It's mighty handy because it transforms the problem into algebra from calculus, which is often simpler to handle.

For our equation \(y^{\backprime \backprime}+y=0\), the characteristic equation is \(r^2+1=0\). This equation is derived by assuming a solution of the form \(e^{rx}\) and plugging it into the homogeneous differential equation. This method, known as the 'exponential guess', simplifies the process of finding the general solution. Solving for \(r\) gives us \(r=\backpm i\), indicating that the solution will involve trigonometric functions like sine and cosine due to the imaginary number.

Homogeneous Differential Equation

In differential equations, 'homogeneous' refers to the absence of any terms without the function we're solving for (in this case, \(y\)) or its derivatives. It means that all terms involve the unknown function or its derivatives. A homogeneous differential equation like the one in our exercise \(y^{\backprime \backprime}+y=0\) always has the trivial solution \(y=0\).

Importance of Homogeneous Solutions

Homogeneous solutions play a critical role in understanding the properties of differential equations. They give us the complementary function, which is the basis of the general solution. By solving the characteristic equation for the given problem, we found that the general solution involves sine and cosine functions, which we'll use along with boundary conditions to find specific solutions.