Consider the BVP $x^2y''-2xy'+2y=0,$ subject to the boundary conditions $$y(1)+\alpha y'(1)=1,$$ $$y(2)+\beta y'(2)=2$$ has a unique solution if
- $\alpha =-1, \beta=2$
- $\alpha =-1, \beta=-2$
- $\alpha =-2, \beta=2$
- $\alpha =-3, \beta=\frac{2}{3}$
Honestly, I have no idea how to solve this problem nor I know any theorem that imposes some conditions on the $\alpha$ and $\beta$ so that the problem has a unique solution. So can anybody push me towards the right theorem or results using which I can solve this problem? Thanks.
Outline:
The differential equation is a Cauchy–Euler equation, which means that it has terms that are of the form $x^ny^{(n)}$. The ansatz for these equations is $y=x^{\alpha}$, and you should find two values of $\alpha$ that work ($1$ and $2$ in this case, so $y=Ax+Bx^2$).
Now substitute into the boundary conditions. For some $\alpha$ and $\beta$, you will find two linearly independent equations for the constants $A$ and $B$, but for others, the equations will just be scalar multiples of one another and won't have a unique solution (i.e. the matrix multiplying $(A,B)$ on the left-hand side will be singular). (I believe in this case you have a unique solution if and only if $2+3b+2ab \neq 0$, according to a quick calculation.)