Construct Green's function to solve the boundary value problem $$ \frac{d^2y}{dx^2} + 4y = f(x), \qquad y(0)=0, \qquad y(\pi/4)=0. $$ Now consider this equation on the interval $[0,L]$ with the same boundary condition $y(0)=0$, $y(L)=0$. For what values of $L$ is this problem impossible? Explain why those values are special.
Someone asked this question in the tutoring center today and it's related to differential equations class. Nobody could solve it. What do they mean by a Green's function? Can we not just solve first the homogeneous $y'' + 4y = 0$ by solving its characteristic equation and then use variation of parameters? I mean that is the way it could be solved, but none of the tutors had a clue as to what is a Green's function
You can solve the equation $y''+4y=f$ subject to $y(0)=y(\pi/4)=0$ using variation of parameters: $$ y = a(x)\sin(2x)+b(x)\sin(2x-\pi/2). $$
The first solution $\sin(2x)$ of $y''+4y=0$ satisfies $y(0)=0$, while the second solution $\sin(2x-\pi/2)$ satisfies $y(\pi/4)=0$. The resulting variation of parameters solution is \begin{align} y(x) =&\frac{1}{2}\sin(2x)\int_{x}^{\pi/4}\sin(2x'-\pi/2)f(x')dx'\\ & +\frac{1}{2}\sin(2x-\pi/2)\int_0^x\sin(2x')f(x')dx'. \end{align} This may also be written as $$ y(x)=\int_0^{\pi/4}G(x,x')f(x')dx, $$ where the Green function $G(x,x')$ is $$ G(x,x')=\left\{\begin{array}{ll} \frac{1}{2}\sin(2x-\pi/2)\sin(2x'),\;\;\; 0 \le x' \le x \\ \frac{1}{2}\sin(2x)\sin(2x'-\pi/2),\;\;\; x \le x' \le\frac{\pi}{4} \end{array} \right. $$ There are ways of interpreting $G(x,x')$ directly, if you consider Dirac delta functions, but variation of parameters will also give you the Green function, as shown above.