I want to check whether following method works in general for finding a solution of the $2^{nd}$ order DE when one solution is given.
$$y''+P(x)y'+Q(x)y=0$$
Let $y_1$ be a solution of the DE and assume that $y_2=v(x)\cdot y_1$ is a solution where $v(x)$ is a function of $x$ and not any constant in $\mathbb{R}$. Hence $y_1$ and $y_2$ will be linearly independent thus making the Wronskian non-zero. Thus, $$\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=\begin{vmatrix} y_1 & v\cdot y_1 \\ y_1' & v\cdot y_1' +v'\cdot y_1 \end{vmatrix}\neq 0\\ \therefore\ v'y_1^2=c_1,\qquad c_1\in \mathbb{R}\\ \therefore\ dv=c_1(y_1^{-2})dx\\ \therefore\ v=c_1\int \frac{dx}{y_1^2}\\ \therefore\ y_2=y_1c_1\int \frac{dx}{y_1^2}\\ \therefore\ y_2=y_1c_1\cdot (I +c_2)$$
So, the only question that remains is finding $c_1\ \&\ c_2$ which can be found by putting some constraints, right?
The Wronskian of this ODE $$y''+Py'+Qy=0~~~~(1)$$ is given as $$y_1y_2'-y_1'y_2=\exp[-\int P dx]$$ Let $y_2(x)=v(x)y_1(x)$, then it follows that $$v'y_1^2=\exp[-\int P dx] \implies v=\int \frac{\exp[-\int {P dx}]}{y_1^2} dx~~~(2)$$ Hence if $y_1$ is one solution of (1), then $y_2(x) = v(x) y_1$ is its other solution. The total (general) solution is given by $y(x)=C_1y_1(x)+C_2 v(x) y_1(x)$, where $C_1, C_2$ are calculated by the additional given particular conditions in a question.
Note that OP's expression for $v(x)$ is incorrect, it should be as given in (2).