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 \(y(x)\).