Question

Determine whether there is any value of the constant \(a\) for which the problem has a solution. Find the solution for each such value. $$ y^{\prime \prime}+4 \pi^{2} y=a+x, \quad y(0)=0, \quad y(1)=0 $$

Short answer

Answer: Yes, there is a value of the constant \(a\). In fact, only \(a=-\frac{1}{4\pi^2}\) results in a solution for the given boundary value problem. The solution is: $$ y(x) = -\frac{1}{2\pi}\sin(2\pi x) + \frac{1}{4\pi^2}x - \frac{1}{4\pi^2} $$

Step-by-step solution

Solve the homogeneous equation

First, we need to solve the homogeneous equation: $$ y'' + 4\pi^2y = 0 $$ This is a second-order linear homogeneous ordinary differential equation with constant coefficients. The characteristic equation is $$ r^2 + 4\pi^2 = 0 $$ Solving for \(r\), we get \(r = \pm 2\pi i\). Since we have complex roots, the general solution of the homogeneous equation is given by: $$ y_h(x) = C_1\cos(2\pi x) + C_2\sin(2\pi x) $$

Find the particular solution

Now, let's find the particular solution of the non-homogeneous equation: $$ y'' + 4\pi^2y = a+x $$ Since the right side is a linear function, we can guess a particular solution of the form \(y_p(x) = Ax + B\). Plugging this into the equation, we get: $$ (Ax + B)'' + 4\pi^2(Ax + B) = a+x $$ This simplifies to \(4\pi^2(Ax + B) = a+x\). Equating the coefficients of \(x\) and the constant terms, we obtain the following system of equations: $$ \begin{cases} 4\pi^2A = 1 \\ 4\pi^2B = a \end{cases} $$ From the first equation, we find that \(A = \frac{1}{4\pi^2}\). Therefore, our particular solution is: $$ y_p(x) = \frac{1}{4\pi^2}x + \frac{a}{4\pi^2} $$

Apply the boundary conditions

Now that we have the general solution, let's consider the boundary conditions: \(y(0) = 0\) and \(y(1) = 0\). The general solution is given by: $$ y(x) = y_h(x) + y_p(x) = C_1\cos(2\pi x) + C_2\sin(2\pi x) + \frac{1}{4\pi^2}x + \frac{a}{4\pi^2} $$ Applying the boundary conditions, we get: $$ \begin{cases} 0 = C_1 + \frac{a}{4\pi^2} \\ 0 = C_1\cos(2\pi) + C_2\sin(2\pi) + \frac{1}{4\pi^2} + \frac{a}{4\pi^2} \end{cases} $$ Since \(\cos(2\pi) = 1\) and \(\sin(2\pi) = 0\), the second equation simplifies to: $$ 0 = C_1 + \frac{1}{4\pi^2} + \frac{a}{4\pi^2} $$

Solve for \(a\) and find the solution

From the system of equations above, we can see that the only possible value for \(a\) is \(- \frac{1}{4\pi^2}\), which makes \(C_1=0\). So now, we can rewrite our solution, keeping in mind this value for \(a\) and \(C_1=0\): $$ y(x) = C_2\sin(2\pi x) + \frac{1}{4\pi^2}x - \frac{1}{4\pi^2} $$ Applying the second boundary condition, \(y(1) = 0\), we find \(C_2 = -\frac{1}{2\pi}\). Finally, our solution is: $$ y(x) = -\frac{1}{2\pi}\sin(2\pi x) + \frac{1}{4\pi^2}x - \frac{1}{4\pi^2} $$

Key concepts

Characteristic Equation

The characteristic equation is a foundational tool in solving second-order linear differential equations with constant coefficients. When faced with a homogeneous equation like \(y'' + 4\pi^2y = 0\), the goal is to find its characteristic equation, which is a simpler algebraic representation of our original differential equation.

To find this equation, we make an assumption that our solution is of an exponential form \(y = e^{rx}\) and substitute back into the homogeneous differential equation. Through this substitution, the differential equations simplify, and we deduce the characteristic equation, \(r^2 + 4\pi^2 = 0\), from which we can solve for the roots \(r\). These roots, either real or complex, dictate the form of the general homogeneous solution, guiding us towards a complete solution of the differential equation.

Homogeneous Equation Solution

Once the characteristic equation has been established, the next step is to determine the solution to the homogeneous equation. Depending on the roots of the characteristic equation, the form of the homogeneous solution, \(y_h\), will vary. In our exercise, the roots turned out to be complex: \(\pm 2\pi i\).

With complex roots, the form of the homogeneous solution is a combination of sine and cosine terms: \(y_h(x) = C_1\cos(2\pi x) + C_2\sin(2\pi x)\). While \(C_1\) and \(C_2\) are arbitrary constants, they will be determined later using boundary conditions to arrive at the specific solution that satisfies the entire differential equation.

Particular Solution Method

The particular solution, \(y_p\), addresses the non-homogeneous part of the equation, which in our case is \(a + x\). A common strategy to find \(y_p\) is to make an educated guess based on the form of the non-homogeneous component, here denoted as the 'method of undetermined coefficients'.

For instance, since our non-homogeneous term is a linear function in \(x\), we predict a particular solution of the form \(y_p(x) = Ax + B\), where \(A\) and \(B\) are constants to be determined. Upon substituting \(y_p\) into the differential equation and comparing coefficients, values for \(A\) and \(B\) are established, leading us to the particular solution that complements the homogeneous solution and reflects the non-homogeneous nature of the original differential equation.

Boundary Conditions

Boundary conditions are essential in defining the specific solution to a differential equation, allowing us to determine the arbitrary constants in the general solution. By applying the given boundary conditions, \(y(0) = 0\) and \(y(1) = 0\), to the general form of the solution, \(y(x) = y_h(x) + y_p(x)\), a system of equations is formulated connecting \(C_1\), \(C_2\), and the constants from the particular solution.

Boundary conditions serve as constraints, and in solving the system of equations arising from these constraints, we pinpoint the exact values for \(C_1\) and \(C_2\). Through this process, the complete specific solution that is consistent with the physical, geometrical, or other conditions imposed by the problem's context is finely sculpted, ensuring that the original differential equation is satisfied under the stipulated conditions.