I've been given a question to solve a fairly straightforward ODE with boundary conditions. However, I've found that I get the trivial solution. Thinking I had done something wrong, I checked my work over and over to no avail. So, I attempted to prove that this must always happen for the given boundary conditions. I will give my proof below, I would appreciate someone looking it over to find any mistakes and providing any insight I may be missing.
Suppose we have the following inhomogenous second-order linear ODE
$$\frac{d^2y}{dx^2}+A(x)\frac{dy}{dx}+B(x)y=S(x)$$
Suppose this has a non-trivial homogeneous solution
$$y_h=C_1y_1+C_2y_2$$
By variation of parameters, we then know the general solution is
$$y=C_1y_1+C_2y_2+\int_a^x\left[y_2(x)\frac{y_1(t)}{W(t)}-y_1(x)\frac{y_2(t)}{W(t)}\right]S(t)\,dt$$
Where $W=y_1y_2'-y_2y_1'$ is the Wronskian. We will assume all of these functions are well-behaved. Now suppose we are given the boundary conditions
$$y(a)=0$$ $$\frac{dy}{dx}(a)=0$$
Substituting this into $y$, we have
$$0=y(a)=C_1y_1(a)+C_2y_2(a)+\int_a^a\left[y_2(a)\frac{y_1(t)}{W(t)}-y_1(a)\frac{y_2(t)}{W(t)}\right]S(t)\,dt$$
So that
$$C_1=-C_2\frac{y_2(a)}{y_1(a)}$$
For the other boundary condition, we calculate $\frac{dy}{dx}$:
$$y'=C_1y_1'+C_2y_2'+y_2'\int_a^x\frac{y_1(t)}{W(t)}S(t)\,dt+y_2\frac{y_1(x)}{W(x)}S(x)-y_1'\int_a^x\frac{y_2(t)}{W(t)}S(t)\,dt-y_1\frac{y_2(x)}{W(x)}S(x)$$
Simplifying yields
$$y'=C_1y_1'+C_2y_2'+\int_a^x\left[y_2'(x)\frac{y_1(t)}{W(t)}-y_1'(x)\frac{y_2(t)}{W(t)}\right]S(t)\,dt$$
Applying the boundary condition gives us
$$0=y'(a)=C_1y_1'(a)+C_2y_2'(a)+\int_a^a\left[y_2'(a)\frac{y_1(t)}{W(t)}-y_1'(a)\frac{y_2(t)}{W(t)}\right]S(t)\,dt$$
So that
$$C_1=-C_2\frac{y_2'(a)}{y_1'(a)}$$
Combining with the other boundary condition yields
$$C_2\left(\frac{y_2(a)}{y_1(a)}-\frac{y_2'(a)}{y_1'(a)}\right)=0$$
So, either $C_2=0$ or $W(a)=0$. Note that $a$ can be any value in the solution's domain, so that $W(a)=0$ implies the solutions are linearly dependent, which cannot be true. Thus the only other possibility is that $C_2=0$, and thus $C_1=0$. This gives us the trivial solution:
$$y=\int_a^x\left[y_2(x)\frac{y_1(t)}{W(t)}-y_1(x)\frac{y_2(t)}{W(t)}\right]S(t)\,dt$$
Which is simply the inhomogenous part of the general solution.
When putting this into physical terms, such as the heat equation, you're saying that the scaled temperature is $0$ at $a$, and there is no heat flux at $a$. Logically, at steady state, this implies that the temperature is $0$ everywhere. I would appreciate any additional insight into this.