Find $\alpha \in \mathbb{R}$ such that there exists a Lie algebra representation of $\mathfrak{sl}_2 (\mathbb{R})$ on $C^\infty(\mathbb{R})$ I am trying to get started with the following exercise in relation to representation theory:

Define differential operators $X, Y, Z_\alpha, \alpha \in \mathbb{R}$, on $C^\infty(\mathbb{R})$ by:
$$X f(t) = f''(t)$$
$$Y f(t) = t^2 f(t)$$
$$Z_\alpha f(t) = t f'(t) + \alpha f(t)$$
For $f \in C^\infty(\mathbb{R})$. Find $\alpha \in \mathbb{R}$ such that there exists a Lie algebra representation of $\mathfrak{sl}_2 (\mathbb{R})$ on $C^\infty(\mathbb{R})$  with image equal to span$_\mathbb{C} \lbrace X , Y , Z_\alpha \rbrace$ (HINT: Find an appropriate basis of  $\mathfrak{sl}_2 (\mathbb{R})$  which is mapped to suitable complex multiples of $X, Y , Z_\alpha$ under the representation)

Even though there is a hint I am having some difficulties getting started with this problem. How do I do this?
 A: We have the operators
$$
Xf = f'', \quad \quad Yf = t^2 f, \quad \quad Z_\alpha f = t f' + \alpha f
$$
Now we compute the three (up to sign) possible commutators. First $X$ and $Y$:
$$ \begin{aligned}
\, [X, Y]f
&= XYf - YXf \\
&= (t^2f)'' - t^2f'' \\
&= t^2 f'' + 4tf' + 2f - t^2f'' \\
&= 4tf' + 2f \\
&= 2 Z_{2} f
\end{aligned} $$
Now $Z_\alpha$ and $X$:
$$ \begin{aligned}
\, [Z_\alpha, X]f
&= Z_\alpha X f - X Z_\alpha f \\
&= tf''' + \alpha f'' - (tf' + \alpha f)'' \\
&= tf''' + \alpha f'' - (2f'' + tf''' + \alpha f'') \\
&= -2f'' \\
&= -2Xf
\end{aligned}$$
and finally $Z_\alpha$ and $Y$:
$$ \begin{aligned}
\,[Z_\alpha, Y]
&= Z_\alpha Y f - Y Z_\alpha f \\
&= t(t^2 f)' + \alpha(t^2 f) - t^2(tf' + \alpha f) \\
&= 2t^2 f + t^3 f' + \alpha t^2 f - t^3 f' - \alpha t^2 f \\
&= 2t^2 f \\
&= 2Yf
\end{aligned} $$
Therefore, setting $Z = -2Z_2$ we have the commutator relations
$$ \begin{aligned}
\,[Y, X] &= Z \\
 [Z, X] &= 2X \\
 [Z, Y] &= -2Y
\end{aligned}$$
which are precisely the commutator relations of $\mathfrak{sl}_2$. If we define $\mathfrak{sl}_2$ to be the three-dimensional Lie algebra with basis $e, f, h$ and Lie bracket
$$ \begin{aligned}
\,[e, f] &= h \\
 [h, e] &= 2e \\
 [h, f] &= -2f
\end{aligned}$$
then the linear map $\varphi$ defined by $\varphi(e) = X$, $\varphi(f) = Y$, and $\varphi(h) = Z$ is a Lie algebra homomorphism, and hence the operators $X, Y, Z$ define a representation of $\mathfrak{sl}_2$.
