Question about possible solutions to a differential equation It is my second week studying differential equations, and I am struggling with the following question:
Which of the following statements are correct?


*

*There is a second order homogenous linear differential equation with continuous coefficients in the open interval $(\frac{π}{2},\frac{3π}{2})$, of which the functions $x$ and $sinx$ are solutions.

*There is a second order homogenous linear differential equation with continuous coefficients in the open interval $(\frac{π}{2},\frac{3π}{2})$, of which the functions $x$ and $x^2$ and $x(x+2)$ are solutions.

*There is a second order homogenous linear differential equation with continuous coefficients in the open interval $(\frac{-π}{2},\frac{π}{2})$, of which the function $x^2$ is a solution.

*There is a second order homogenous linear differential equation with constant coefficients in the open interval $(\frac{π}{2},\frac{3π}{2})$, of which the functions $x$ and $x^2$ and $x(x+2)$ are solutions.

*There is a second order non-homogenous linear differential equation with continuous coefficients in the open interval $(0,2)$, to which the functions $x$ and $x^2$ and $x^3$ are solutions.
I am pretty sure that 1. and 4. are incorrect, but am struggling with the rest. Could anybody please advise?
Thank you!
 A: Simple helpful facts: 


*

*i) At a regular point (coefficients all continuous) of an explicit homogeneous second order ODE $$y''+py'+qy=0,$$ if both $y(x_0)=0$ and $y'(x_0)=0$, then $y=0$ on the whole interval of coefficient continuity.

*ii) Given any two functions $y_1$, $y_2$, you get a homogeneous second order ODE having them as solutions as $$0=\det\pmatrix{y_1&y_2&y\\y_1'&y_2'&y'\\y_1''&y_2''&y''}.$$ Singular points are the roots of the leading coefficient $y_1y_2'-y_1'y_2$.

*iii) Given any three functions $y_1$, $y_2$, $y_3$, you get a  second order ODE having them as solutions as $$0=\det\pmatrix{1&1&1&1\\y_1&y_2&y_3&y\\y_1'&y_2'&y_3'&y'\\y_1''&y_2''&y_3''&y''}.$$ Singular points are again the roots of the leading coefficient $$\det\pmatrix{1&1&1\\y_1&y_2&y_3\\y_1'&y_2'&y_3'}.$$

*iv) If an explicit homogeneous linear ODE has constant coefficients, then it is completely determined by the roots of its characteristic polynomial. If a solution contains a term $p(x)e^{\lambda x}$, then $λ$ is a root of multiplicity at least $1+\deg p$.



*

*By ii), the only singular points of the ODE for these functions are the roots of $x\cos x-\sin x$, of which there is one in the interval by the intermediate value theorem.

*The third function is a linear combination of the first two, and the leading coefficient in ii) is $2x^2-x^2=x^2$, so that only $x=0$ has to be excluded from the domain.

*See i)

*By iv), $0$ needs to be characteristic root of multiplicity at least 3. Which is impossible.

*By iii), such an ODE exists. The given interval can be a domain if the leading coefficient 
\begin{align}
\det\pmatrix{1&1&1\\x&x^2&x^3\\1&2x&3x^2}
&=x\det\pmatrix{1&1&1\\1&x&x^2\\1&2x&3x^2}
=x\det\pmatrix{1&1&1\\0&x-1&x^2-1\\0&x&2x^2}
\\[1em]
&=x^2(x-1)\det\pmatrix{1&x+1\\1&2x}
=x^2(x-1)^2
\end{align} 
has no roots in it, which is obviously not the case.
