Differential Equations Constant 
The function $y(x)$ satisfies the linear equation $$y'' + p(x)y' + q(x)y = 0.$$
The Wronskian $W(x)$ of two independent solutions, denoted $y_1(x)$ and $y_2(x)$, is defined to be $$W(x) = \left|\begin{array}{ccc} y_1 & y_2 \\ y_1' & y_2' \end{array}  \right|.$$
Let $y_1(x)$ be given. Use the Wronskian to determine a first-order inhomogeneous differential equation for $y_2(x)$. Hence, show that $$y_2(x) = y_1(x) \int_{x_0}^x \frac {W(t)}{y_1(t)^2} \mathrm d t . $$

I get by expanding the determinant and using an integrating factor $$ \frac {y_2(x)} {y_1(x)} =  \int \frac {W(x)}{y_1(x)^2} \mathrm d x $$
hence
$$ \frac {y_2(x)} {y_1(x)} -  \frac {y_2(x_0)} {y_1(x_0)} =  \int_{x_0}^x \frac {W(t)}{y_1(t)^2} \mathrm d t . $$
But why can I just decide $y_2(x_0)=0$ if $x_0$ is arbitrary? And if it's for a specific value of $x_0$ (the question is rather vague about this...) then how do I know that in general there exists $x_0$ such that $y_2(x_0)=0$?
 A: As in the method of variation of parameters, if $y_1(x)$ is a known solution of the given ODE, we will seek a second linearly independent solution of the form $y_2(x)=v(x)y_1(x)$ where $v(x)$ is a function that is to be determined.
Suppressing the $x$ dependence for notational convenience:
\begin{align}
y_2&=vy_1\\
y_2'&=v'y_1+vy_1'\\
y_2''&=v''y_1+v'y_1'+v'y_1'+vy_1''=v''y_1+2v'y_1'+vy_1''
\end{align}
and substituting these into the ODE,
\begin{align}
y_2''+py_2'+qy_2=v''y_1+2v'y_1'+vy_1''+p(v'y_1+vy_1')+q(vy_1)&=0\\
v\underbrace{(y_1''+py_1'+qy_1)}_{=0}+v''y_1+(2y_1'+py_1)v'&=0\\
v''+\left({2y_1'\over y_1}+p\right)v'&=0.\tag{1}
\end{align}
Let $w=v'$ so $w'=v''$ (this is called reduction of order), $(1)$ reduces to the first linear order equation (that the exercise asked for) in $w$:
$$
w'+\left({2y_1'\over y_1}+p\right)w=0.\tag{2}
$$
The integrating factor is $$m(x)=\exp\left(\int \left({2y_1'\over y_1}+p\right) dx\right)=\exp(2\ln|y_1|)\exp\left({\int p\,dx}\right)=y_1^2\exp\left({\int p\,dx}\right),$$
so multiplying $(2)$ by $m(x)$,
$$
[m w]'=0\implies w={C\over y_1^2\,e^{\int p\,dx}}={Ce^{-\int p\,dx}\over y_1^2},
$$
and thus
$$
v=\int {Ce^{-\int p\,dx}\over y_1^2}\,dx. \tag{3}
$$
To reconcile this with your desired result, appeal to Abel's Identity which says
$$
W(y_1,y_2)(x)=C\exp\left(-\int p\,dx\right).
$$
Then $(3)$ becomes
$$
v=\int {W(y_1,y_2)\over y_1^2}\,dx,
$$
and this in turn reveals the (form of the) second linearly independent solution we sought:
$$y_2(x)=v(x)y_1(x)=y_1(x)\int {W(y_1,y_2)(x)\over y_1^2(x)}\,dx.
$$
PS I don't know if you have Abel's Identity at your disposal or if that was intended to be part of the exercise. If the latter, the link above provides the proof of that.
A: Your solution is correct, but it does not really matter for the application of the method as long as you apply the initial conditions in the end.
Using this method you are given a solution $y_1$ and want to use this to find another linear independent solution $y_2$ and then you know that the general solution to the ODE reads
$$y = Ay_1(x) + By_2(x)$$
for some constants $A,B$ that are determined by the initial conditions. Now as you have shown
$$y_2 = y_1\int \frac{W(t)}{y_1^2(t)}dt$$
so in general we have
$$y_2 = y_1\int_{x_0}^{x} \frac{W(t)}{y_1^2(t)}dt + Cy_1$$
for some (integration constant) $C$ where you can fix $C$ in terms of $x_0$ as in your solution by plugging in $x=x_0$ above and solving for $C$. However, note that we can, without loss of generality, just put $C=0$ here since the expression we obtain for $y$ (which is the thing we are after) reads
$$y = Ay_1 + By_2 = A'y_1 + By_1\int_{x_0}^{x} \frac{W(t)}{y_1^2(t)}dt$$
where $A'=(A+BC)$ is just another constant which we will solve for directly (not caring about the value of $C$) when we apply the initial conditions. So to conclude: if you apply the initial conditions in the end you can in fact simply define $y_2 = y_1\int_{x_0}^{x}\frac{W(t)}{y_1^2(t)}dt$ for any $x_0$ you want and the method will give the correct result.
