Find the solutions to this ODE for an arbitrary $\lambda$ 
For $\lambda \in \mathbb{R}$, consider the boundary value problem 
$$x^2\frac{d^2y}{dx^2}+2x\frac{dy}{dx}+\lambda y=0, \quad x\in [1,2], \qquad y(1)=y(2)=0$$
Which of the following statements is true?
  
  
*
  
*there exist a $\lambda_0 \in \mathbb{R}$ such that problem $P_{\lambda}$ has a non trivial solution for $\lambda>\lambda_0$.
  
*$\{\lambda \in \mathbb{R}: P_{\lambda}$ has a non trivial solution$\}$ is a dense subset of $\mathbb{R}$.
  
*For any continuous function $f:[1,2] \to \mathbb{R}$ with $f(x)\neq 0$ for some $x \in [1,2]$ there exist a solution u of Problem for some
  $\lambda \in \mathbb{R}$ such that $\int\limits_1 ^2 fu \neq 0$
  
*there exist a $\lambda \in \mathbb{R}$ such that Problem $P_{\lambda}$  has two linearly independent solutions.

As- c
I know (4) is incorrect as Wronskian of the system comes out 0. But how to look for others? 
 A: Solve this second-order linear ordinary differential equation:
$$x^2y''(x)+2y'(x)+\lambda y(x)=0\Longleftrightarrow$$

Assume the solution will be proportional to: $x^\mu$ for some constant $\mu$.
Substitute $y(x)=x^\mu$:

$$x^2\cdot\frac{\text{d}^2}{\text{d}x^2}\left(x^\mu\right)+2\cdot\frac{\text{d}}{\text{d}x}\left(x^\mu\right)+\lambda x^\mu=0\Longleftrightarrow$$

Substitute:


*

*$$\frac{\text{d}}{\text{d}x}\left(x^\mu\right)=\mu x^{\mu-1}$$

*$$\frac{\text{d}^2}{\text{d}x^2}\left(x^\mu\right)=(\mu-1)\mu x^{\mu-2}$$


$$x^2\cdot\left((\mu-1)\mu x^{\mu-2}\right)+2\cdot\left(\mu x^{\mu-1}\right)+\lambda x^\mu=0\Longleftrightarrow$$
$$\mu^2x^\mu+\mu x^\mu+\lambda x^\mu=0\Longleftrightarrow$$
$$x^\mu\left(\mu^2+\mu+\lambda\right)=0\Longleftrightarrow$$

Assuming $x=\ne0$, the zeros must come from the polynomial:

$$\mu^2+\mu+\lambda=0\Longleftrightarrow$$
$$\mu=\frac{-1\pm\sqrt{1-4\lambda}}{2}$$
So, as solution we get:
$$y(x)=\text{C}_1x^{\frac{\sqrt{1-4\lambda}-1}{2}}+\text{C}_2x^{-\frac{1+\sqrt{1-4\lambda}}{2}}$$
Now, we need to solve for the initial conditions $y(1)=y(2)=0$:
$$
\begin{cases}
\text{C}_1\cdot1^{\frac{\sqrt{1-4\lambda}-1}{2}}+\text{C}_2\cdot1^{-\frac{1+\sqrt{1-4\lambda}}{2}}=0\\
\text{C}_1\cdot2^{\frac{\sqrt{1-4\lambda}-1}{2}}+\text{C}_2\cdot2^{-\frac{1+\sqrt{1-4\lambda}}{2}}=0
\end{cases}\Longleftrightarrow
\begin{cases}
\text{C}_1+\text{C}_2=0\\
\text{C}_1\cdot2^{\frac{\sqrt{1-4\lambda}-1}{2}}+\text{C}_2\cdot2^{-\frac{1+\sqrt{1-4\lambda}}{2}}=0
\end{cases}
$$
If we solve the system we get that (real solution):
$$\text{C}_2=-\text{C}_1,\lambda=\frac{1}{4}$$
