Find the first $4$ Hermite polynomials using a recursion relation Given the Probabilists' Hermite differential equation: $$U''-xU'+\lambda U=0\tag{1}$$
A book question asks me to: 

Find the first $4$ polynomial solutions (for $\lambda = 0,1,2,3$), each normalised such that the highest power of $x$ has a coefficient of unity.


So I substituted $$U=\sum_{n=0}^{\lambda}C_nx^n\tag{2}$$ into $(1)$ and equated coefficients of $x^n$ to get the recursion relation:
$$C_{n+2}=\frac{(n-\lambda)}{(n+1)(n+2)}C_n$$
So I am looking to find the values of the coefficients $C_0,C_1,C_2$ for $$U=\sum_{n=0}^{3}C_nx^n=C_0+C_1x+C_2x^2+x^3$$
where we are given that $\color{blue}{C_3}=1$.
Using the recursion relation I find that for $(n=0, \lambda=2)$: $$C_2=-C_0$$ 
and for $(n=1, \lambda=3)$: $$\color{blue}{C_3}=-\frac13C_1\implies C_1=-3$$
This is as far as I can get to in this question.

The book answer simply states:

Using the recurrence relations, and setting the coefficient of the
  highest power to $1$, $U_0=1, \,U_1=x,\, U_2=x^2-1\,$ and $U_3=x^3-3x$

I know from this page on Wikipedia that the book answer is right, but the answer is not very helpful to me as I have no idea why $U_0=1$. Does this mean that $C_0=1$, if so how did the author deduce this from the recursion relation? I showed that $C_1$ is equal to $-3$. So why is $U_1\ne-3x$? The same misunderstanding follows for $U_2$ and $U_3$. In fact I don't understand why $U_3$ doesn't contain an $x^2$ term and a constant term. Also, where is the $x$ term in $U_2$ and the constant term in $U_1$? 
I have only just started reading about Hermite polynomials so my understanding is very weak (apologies). 
Is there any chance someone could explain as simply as possible how to obtain $U_0,U_1,U_2,U_3$? 
Thank you.
 A: You should add a second index to the recurrence formula in order to avoid confusion. 
$$C_{n+2}^{(\lambda)}=\frac{(n-\lambda)}{(n+1)(n+2)}C_n^{(\lambda)}$$
With this, it is an immediate consequence that in even indexed polynomial $U_0, U_2,\dots$ the odd coefficients $C_{1}^{(\lambda)}, C_{3}^{(\lambda)}$ are zero and analogous for the odd indexed polynomials. So you have the trivial cases
$$U_0(x)=1$$ 
$$U_1(x)=x$$ 
For $U_2(x) = C_{0}^{(2)} + x^2$ you get 
$$1=C_{2}^{(2)} = \frac{(0-2)}{(0+1)(0+2)}C_{0}^{(2)}$$
i.e. $\color{red}{C_{0}^{(2)}}=-1$ and $U_2(x)= x^2-1$.
For $U_3(x) = C_{1}^{(2)}x + x^3$ you get 
$$1=\color{red}{C_{3}^{(3)}} = \frac{(1-3)}{(1+1)(1+2)}C_{1}^{(3)}=-\frac{1}{3}C_{1}^{(3)}$$
i.e. $\color{red}{C_{1}^{(3)}}=-3$ and $U_2(x)= x^3-3x$.
A: $$U''-xU'+\lambda U=0$$
This is more a long comment than an answer. If fact I only answer to  OP question : "I have no idea why $U_0 =1$. Does this mean that $C_0 =1$ ? " : See below the case $\lambda=0$.
The answer with the method of series and recurrence relations has already been given. No need to repeat it. Of course, it isn't a smart method that is  used below.
Case $\lambda=0 \quad\to\quad U''-xU'=0$
The polynomial solution comes from $U'=0 \quad\to\quad U=c=$constant.
This is the highest power of $x$ i.e.: $x^0$ . Taking account of the request of normalisation $c=1$
$$U_0(x)=1$$
Case $\lambda=1$ 
$U''-xU'+U=0\quad $Polynomial solution : $U=ax+b$
$(ax+b)''-x(ax+b)'+(ax+b)=0 \quad\to\quad \begin{cases} \text{any }a \\ b=0\end{cases} \quad\to\quad U=ax$
$a=1$ to be normalized.
$$U_1(x)=x$$
Case $\lambda=2$
$U''-xU'+2U=0\quad$ Polynomial solution $U=ax^2+bx+c$
$2a-x(2ax+b)+2(ax^2+bx+c)=0 \quad\to\quad \begin{cases} \text{any }a \\ b=0\\ c=-a\end{cases} \quad\to\quad U=ax^2-a$
$a=1$ to be normalized.
$$U_2(x)=x^2-1$$
Case $\lambda=3$
$U''-xU'+3U=0\quad$ Polynomial solution $U=ax^3+bx^2+cx+d$
$(6ax+2b)-x(3ax^2+2bx+c)+3(ax^3+bx^2+cx+d)=0$
$\begin{cases} \text{any }a \\b=0 \\ c=-3a \\ d=0\end{cases} \quad\to\quad U=ax^3-3a \qquad a=1$ to be normalized.
$$U_3(x)=x^3-3x$$
Case $\lambda=4$
$U''-xU'+4U=0\quad$ Polynomial solution $U=ax^4+bx^3+cx^2+dx+E$
$(12ax^2+6bx+2c)-x(4ax^3+3bx^2+2cx+d)+4(ax^4+bx^3+cx^2+dx+E)=0$
$\begin{cases} \text{any }a \\b=0 \\ c=-6a \\ d=0 \\ E=3a\end{cases} \quad\to\quad U=ax^4-6ax^2+3a \qquad a=1$ to be normalized.
$$U_4(x)=x^4-6x^2+3$$
