I have found two computationally similar yet conceptually dissimilar methods for applying reduction of order to an ODE. The first starts with $y_{general} = vy_1$, while the second starts with $y_2 = vy_1$. From there, the two more or less converge until the end; plug $vy_1$ into the ODE, do straightforward algebra, and make the obvious substitutions as needed. The second method yields a second particular solution $y_2$, so from there the general solution must be constructed as $y_{general} = Cy_1 + Dy_2$. However, the first method yields a general solution directly, without this final step. Let's try the example $t^2 y'' + ty' - y = 0, y_1 = t$.
Method 1
$y_{general} = vt \implies t^2(v''t + 2v') + t(v't + v) - vt = 0 \implies v' = At^{-3} \implies v = At^{-2} + B \implies y_{general} = (At^{-2} + B)t = At^{-1} + Bt$
Method 2
$y_2 = vt \implies t^2(v''t + 2v') + t(v't + v) - vt = 0 \implies v' = At^{-3}$, but our goal is a particular solution, so we'll choose $A = 1$ and thus $v' = t^{-3} \implies v = t^{-2} + B$, but our goal is a particular solution, so we'll choose $B = 0$ and thus $v = t^{-2} \implies y_2 = (t^{-2})t = t^{-1} \implies y_{general} = Cy_1 + Dy_2 = Ct + Dt^{-1}$
$At^{-1} + Bt$ and $Ct + Dt^{-1}$ are equal. We have arrived at the same general solution via both methods, but I find the computational-similarity-yet-conceptual-dissimilarity between the two quite suspicious. Why do these two identically named methods always arrive at the same result, if in fact they do.
We are given
$$\tag 1 t^2 y'' + ty' - y = 0, y_1 = t$$
This is a homogeneous linear equation and we can use many methods to find a general solution.
For example, we can solve this problem as a homogeneous linear equation, exact equation or use a substitution method like Reduction of Order.
Method I: Solve as an Euler-Cauchy type equation.
Assume $$y = t^m \implies y' = m t^{m-1} \implies y'' = m(m-1) t^{m-2}$$
Substituting into $(1)$, we find that $m^2 - 1 = 0 \implies m = \pm~ 1$, which leads to the general solution $$y = c_1 t + \dfrac{c_2}{t}$$
Method II: Solve using Reduction of Order
We are told that a solution is $y_1 = t$.
We find the second solution using
$$y_2 = v t \implies y_2' = v' t + v, y_2'' = v'' t + 2 v'$$
Substituting into $(1)$
$$t^3 v'' + 3 t^2 v' = 0$$
Let $w = v' \implies w' = v''$, which leads to a separable equation and find $w = \dfrac{c_1}{t^3}$.
We now have
$$v' = w = \dfrac{c_1}{t^3} \implies v = \dfrac{c_1}{t^2}$$
Dropping the constants, we have two solutions as $y_1 = t, y_2 = v t = \dfrac{1}{t}$.
We can now write the general solution as
$$y = y_1 + y_2 = c_1 t + \dfrac{c_2}{t}$$
Compare that to Method I. This is precisely what was done in the Youtube Video.
You can also try the Exact-Equation approach as a third option.
There are not two different Reduction of Order methods.
Update
In the video, we are given
$$\tag 2 x^2 y'' + 5 x y' - 5 y = 0, y_1 = x$$
Let $y_2 = v x \implies y_2' = v' x + v \implies y_2'' = v'' x + 2 v'$
Substituting into $(2)$
$$x^3 v'' + 7 x^2 v' = 0$$
Let $w = v' \implies w' = v''$, so
$$x^3 w' + 7x^2 w = 0$$
This is Separable Equation and we get $w = \dfrac{c_1}{x^7}$.
We now have $v' = \dfrac{c_1}{x^7}$ and anti-differentiating, we get
$$v = \dfrac{c_1}{x^6} + c_2$$
This is where the video deviates. What have two choices we can make. We can $1)$, choose $c_2$ to be zero or $2)$ keep $c_2$.
Case 1: If we choose $c_2 = 0$, we get
$$y_2 = v x = \left(\dfrac{c_1}{x^6}\right)x = \dfrac{c_1}{x^5}$$
the general solution in this case is
$$y = y_1 + y_2 = c_1 x + \dfrac{c_2}{x^5}$$
Case 2: If we choose $c_2 \ne 0$, we get the general solution
$$y = v x = \left(\dfrac{c_1}{x^6} + c_2\right)x = \dfrac{c_1}{x^5} + c_2 x$$
Additionally, I think you should work through a different example and these using both methods.