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. $$ y^{\prime \prime}+\pi^{2} y=a+x, \quad y(0)=0, \quad y(1)=0 $$

Short answer

The value of \(a\) for which the problem admits a solution is \(a=-1\). The corresponding specific solution to the initial value problem is given by: $$ y(x)=\frac{1}{\pi^2}\cos(\pi x)+C_2\sin(\pi x)+\frac{1}{\pi^2}x-\frac{1}{\pi^2}, $$ where \(C_2\) can be any real number.

Step-by-step solution

Find the complementary solution

First, we will find the complementary function by solving the homogeneous version of the ODE given \(y''+\pi^2y=0\). This is a second-order linear homogeneous ODE with constant coefficients, the characteristic equation is: $$ r^2+\pi^2=0 $$ Solving the above equation for \(r\), we get complex conjugate roots \(r=\pm\pi i\). So, the complementary function is given by: $$ y_c(x)=C_1\cos(\pi x)+C_2\sin(\pi x) $$ where \(C_1\) and \(C_2\) are constants.

Find the particular solution

Now, we will find the particular solution for the inhomogeneous ODE. Since the non-homogeneous term is a linear function of \(x\), we will try a particular solution of the form: $$ y_p(x)=Ax+B $$ where \(A\) and \(B\) are constants that we need to determine. Taking the first and second derivatives of \(y_p(x)\), we get: $$ y_p'(x)=A, $$ and $$ y_p''(x)=0 $$ Now, substitute \(y_p(x)\) and its derivatives into the original inhomogeneous ODE: $$ 0+\pi^2(Ax+B)=a+x $$ Comparing coefficients, we have: $$ \begin{cases} \pi^2A=1, \\ \pi^2B=a. \end{cases} $$ Solving for \(A\) and \(B\), we get $$ A=\frac{1}{\pi^2},\quad B=\frac{a}{\pi^2} $$ Thus, the particular solution is $$ y_p(x)=\frac{1}{\pi^2}x+\frac{a}{\pi^2}. $$

Combine complementary and particular solutions

The general solution for \(y(x)\) is the sum of the complementary and particular solutions: $$ y(x)=y_c(x)+y_p(x)=C_1\cos(\pi x)+C_2\sin(\pi x)+\frac{1}{\pi^2}x+\frac{a}{\pi^2} $$

Apply initial conditions

Now, we will apply the given initial conditions \(y(0)=0\) and \(y(1)=0\). For the initial condition \(y(0)=0\), we have $$ 0=C_1\cos(0)+C_2\sin(0)+\frac{1}{\pi^2}(0)+\frac{a}{\pi^2} $$ This simplifies to $$ C_1+\frac{a}{\pi^2}=0 $$ For the initial condition \(y(1)=0\), we have $$ 0=C_1\cos(\pi)+C_2\sin(\pi)+\frac{1}{\pi^2}(1)+\frac{a}{\pi^2} $$ This simplifies to $$ -C_1+\frac{1}{\pi^2}+\frac{a}{\pi^2}=0 $$

Determine the value of constant \(a\)

Solving the equations obtained from the initial conditions, we get $$ \begin{cases} C_1+\frac{a}{\pi^2}=0, \\ -C_1+\frac{1}{\pi^2}+\frac{a}{\pi^2}=0. \end{cases} $$ Adding both equations, we get $$ \frac{a+1}{\pi^2}=0 $$ Solving for \(a\), we obtain the unique value of \(a\) for which the given problem has a solution: $$ a=-1 $$

Find the final solution

Substitute the found value of \(a\) into the general solution of \(y(x)\): $$ y(x)=C_1\cos(\pi x)+C_2\sin(\pi x)+\frac{1}{\pi^2}x-\frac{1}{\pi^2} $$ Now, we can use the initial condition \(y(0)=0\) to find the constant \(C_1\): $$ 0=C_1-\frac{1}{\pi^2} $$ Solving for \(C_1\), we have $$ C_1=\frac{1}{\pi^2} $$ Thus, the solution for \(y(x)\) and the given value of \(a=-1\) is $$ y(x)=\frac{1}{\pi^2}\cos(\pi x)+C_2\sin(\pi x)+\frac{1}{\pi^2}x-\frac{1}{\pi^2} $$ Since we already satisfied both initial conditions, the constant \(C_2\) can be any real number, giving a family of solutions for \(y(x)\).

Key concepts

Boundary Value Problems

Boundary Value Problems (BVPs) are a fascinating area of differential equations. These problems involve finding a solution to a differential equation that satisfies certain conditions at the boundaries of the interval.
For instance, in our example, we have the conditions \( y(0) = 0 \) and \( y(1) = 0 \), together with the differential equation \( y'' + \pi^2 y = a + x \).

Such problems are common in physical applications where you need to determine unknown functions based on parameters at the surface or outer limits of a domain, such as temperature distribution in a rod or vibrations in a bridge.
Unlike initial value problems (IVPs), BVPs can be more complex to solve because their solutions are influenced by conditions at two different points! It's essential to find a solution that fits all given boundaries, which often requires blending complementary and particular solutions.

Second-order Linear ODE

Second-order Linear ODEs are ordinary differential equations where the highest derivative is the second derivative.
In our example, the equation \( y'' + \pi^2 y = a + x \) is a second-order ODE. Here, \( y'' \) represents the second derivative of \( y \) with respect to \( x \).

These equations can be challenging but powerful as they appear in many applications, such as modeling the motion of waves and the dynamics of structures under varying forces.
Second-order linear ODEs are typically expressed as \( ay'' + by' + cy = f(x) \), where \( a \), \( b \), and \( c \) are constants.
When analyzing such equations, it's vital to identify whether they are homogeneous or not, as this changes the approach for finding solutions.

Particular Solution

A particular solution is a specific solution to a non-homogeneous differential equation.
It includes the terms from the non-homogeneous part, in our case, \( a + x \). The goal is to find a single function that satisfies the differential equation.

To find the particular solution, we often choose a trial solution that resembles the non-homogeneous term.
Here, our trial solution is of the form \( y_p(x) = Ax + B \).

• Different forms for different trial solutions exist depending on the structure of \( f(x) \).

By substituting \( y_p(x) \) into the original equation, we solve for the constants to fit the equation, leading us to \( y_p(x) = \frac{1}{\pi^2}x + \frac{a}{\pi^2} \).

This solution uniquely addresses the conditions imposed by the non-homogenic part but doesn't yet consider the boundary conditions of the problem.

Homogeneous Equation

The homogeneous equation derived from our original problem is \( y'' + \pi^2 y = 0 \). This equation results by removing the non-homogeneous part, making it crucial to determine the structure of solutions for the ODE.

Solving homogeneous equations often starts with finding the characteristic equation, \( r^2 + \pi^2 = 0 \).
The roots of this equation, here \( r = \pm \pi i \), are complex conjugates and lead to solutions in terms of sine and cosine functions.

• The resulting complementary function is \( y_c(x) = C_1\cos(\pi x) + C_2\sin(\pi x) \).

This part of our solution accounts for the response solely due to the homogeneous system.
It's the backbone upon which the particular solution builds to form the general solution.

The homogeneous equation gives flexibility that allows us to adjust constants \( C_1 \) and \( C_2 \) to satisfy boundary conditions, making it key to unreachable solutions when alone.