For each positive integer $m$, find a polynomial $y(x)$ such that $xy''(x)+(1-x)y'(x)+my(x)=0$ 
Show that, for every $m \in \mathbb{N}$, there is a polynomial $y(x)$ such that $$\bigg(xD^2+(1-x)D+m\bigg)y(x)=0$$  

This is a past year exam question in my ODE module and my lecturer did not provide any solutions for this particular paper. I'd still like to know how can i go about proving this question at my own pace. I have completely no clue as to how to start this question so any help would be much appreciated!
 A: $\require{cancel}$
Hoping all my calculations are correct, this might be an answer:
Let: $$y(x)=\sum_{n=0}^\infty C_nx^{n+r},\quad y'(x)=\sum_{n=0}^\infty C_n(n+r)x^{n+r-1},\quad y''(x)=\sum_{n=0}^\infty C_n(n+r)(n+r-1)x^{n+r-2}$$
Substituting in the original equation we obtain:
$$\sum_{n=0}^\infty C_n(n+r)(n+r-1)x^{n+r-1}+\sum_{n=0}^\infty C_n(n+r)x^{n+r-1}-\sum_{n=0}^\infty C_n(n+r)x^{n+r}+\\+\sum_{n=0}^\infty C_nm\cdot x^{n+r}=0$$
Factoring out $x^r$ and making the simple substitution $n=k+1$ for the first two sums and $n=k$ for the other two, we obtain:
$$x^r\left[\sum_{k=-1}^\infty C_{k+1}(k+r+1)(k+r)x^{k}+\sum_{k=-1}^\infty C_{k+1}(k+r+1)x^{k}-\sum_{k=0}^\infty C_k(k+r)x^{k}+\sum_{k=0}^\infty C_km\cdot x^{k}\right]=0$$
Evaluating the first two sums for $k=-1$ and putting them together, leads to:
$$(r^2-r)C_0x^{-1}+rC_0x^{-1}+\\+\sum_{k=0}^{\infty}\left[C_{k+1}(k+r+1)(k+r)+C_{k+1}(k+r+1)-C_k(k+r)+mC_k\right]x^k=0$$ 
From the above equation we are able to obtain the indicial polynomial and the recurrence relation for $k=\{0,1,2,3...\}$:
$$ \begin{align*}
        (r^2-r)C_0x^{-1}+rC_0x^{-1}&= 0 &C_{k+1}(k+r+1)(k+r)+C_{k+1}(k+r+1)-C_k(k+r)+mC_k=0\\
r&= 0 &C_{k+1}=\frac{(k+r-m)C_k}{k^2+2rk+2k+r^2+r+1}
        \end{align*}$$
Subsituting $r$ in the recurrence relation, we obtain:
$$C_{k+1}=\frac{(k-m)C_k}{k^2+2k+1}$$
Evaluating for different values of $k$,
\begin{align}
\text{For } k=0:& \quad\quad C_1 =-mC_0\\
\text{For } k=1:& \quad\quad C_2 =\frac{1-m}4C_1\\
\text{For } k=2:& \quad\quad C_3 =\frac{2-m}9C_2=\frac{(2-m)(1-m)}{36}C_1\\
\text{For } k=3:& \quad\quad C_4 =\frac{3-m}{16}C_3=\frac{(3-m)(2-m)(1-m)}{576}C_1\\
\vdots
\end{align}
Therefore, knowing that:
$$y(x)=\sum_{n=0}^\infty C_nx^{n+r}$$
and with all the information obatined, $y(x)$ can be expressed as:
$$y(x)=\sum_{n=0}^\infty C_nx^{n}=C_0-mC_0x+\frac{1-m}4C_1x^2+\frac{(2-m)(1-m)}{36}C_1x^3+\frac{(3-m)(2-m)(1-m)}{576}C_1x^4+\cdots$$
Thus:
$$y_1(x)=C_0(1-mx)$$ $$ y_2(x)=C_1\left(\frac{1-m}4x^2+\frac{(2-m)(1-m)}{36}x^3+\frac{(3-m)(2-m)(1-m)}{576}x^4+\cdots\right)$$
As we can see from $y_2(x)$, if $m\in\mathbb{N}$ the differential equation will have as a solution a polynomial of degree $m$. 

For instance, let $m=3$, then:
$$y(x)=C_0-3C_0x-\frac12C_1x^2+\frac1{18}C_1x^3+\cancel{0\cdot C_1x^4}+\cancel{0\cdot C_1x^5}+\cancel{0\cdot C_1x^6}\cdots$$
As you can see, the solution of the differential equation is a polynomial with highest degree equal to $3=m$.
