Finding Second Solution for Hermite Differential Equation through reduction of order One can use the ordinary power series solution to find one solution of the Hermite Differential Equation
$$ y''(x) - 2 x y'(x) + \lambda y(x) = 0$$
Can one use the reduction of order technique to find another linearly independent solution to the equation, of the form $ y_2 (x) = v(x) y_1 (x) $ ?
 A: Original post: $y''(x)-2xy'(x) + \lambda x=0$ (not an Hermite's equation)
This differential equation can be easily solved directly. Let $z(x) := y'(x)$. Therefore
$$
z' -x(2z-\lambda) = 0 \Rightarrow z(x) = c_1e^{x^2}+\frac{\lambda}{2}.
$$
Since $z(x) = y'(x)$,
$$
\boxed{y(x) = c_1\mathrm{erfi}(x) + \frac{\lambda x}{2} + c_2,}
$$
where $\mathrm{erfi}(x)$ is the imaginary error function defined as
$$
\mathrm{erfi}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{t^2}\mathrm{d}t.
$$
If you want to use the reduction of order technique, we begin by solving the homogeneous equation
$$
z_1'=2xz_1 \Rightarrow z_1(x) = c_1e^{x^2}.
$$
Now, we consider a second solution written as $z_2(x)=v(x)z_1(x) \Rightarrow z_2'(x) = v'(x)z_1(x) + 2xv(x)z_1(x)$. Therefore,
$$
\begin{aligned}
v'(x)&z_1(x) + 2xv(x)z_1(x) -x(2v(x)z_1(x) -\lambda) = v'(x)z_1(x) + \lambda x = 0 \\
&\Rightarrow v'(x) = -\frac{\lambda}{c_1} x e^{-x^2} \Rightarrow v(x) = \frac{\lambda}{2c_1}e^{-x^2} \Rightarrow z_2(x) = \frac{\lambda}{2}.
\end{aligned}
$$
Finally, $z(x) = z_1(x) + z_2(x) = c_1e^{x^2} + \lambda/2 = y'(x)$, so
$$
\boxed{y(x) = c_1\mathrm{erfi}(x) + \frac{\lambda x}{2} + c_2,}
$$
as expected.
Edited post: $y''(x)-2xy'(x) + \lambda y(x)=0$ (Hermite's equation)
The general solution to this one is
$$
y(x) = c_1H_{\lambda/2}(x) + c_2\,{_1F}_1\left(-\frac{\lambda}{4};\frac{1}{2};x^2\right),
$$
being $H_n(x)$ the $n$-th Hermite polynomial and ${_1}F_1(a; b; x)$ the confluent hypergeometric function of the first kind. This solution can be simplified for certain values of $\lambda$. Of course you can use the reduction of order technique as showed but it doesn't seem to be a useful method.
