Solution of second ordinary equation i have the following question. Let $\phi_1$ and $\phi_2$ fundamental system solutions on an interval $I$ for the second order equation 
$$
y''+a(x)y= 0.
$$
Prove that there exists fundamental system solutions $\{y_1,y_2\}$ such that the Wronksian $W[y_1,y_2]$ satisfies $W[y_1,y_2]=1$.
So, I know that $\{y_1,y_2\}$ is a system of fundamental solution means that any solution $y$ of edo is written: $y(x)= c_1 y_1(x)+ c_2 y_2(x)$ where $c_1$ and $c_2$ arbitrary contacts. Then $W[y_1,y_2]= y_1(x)y'_2(x)-y_2(x)y_1'(x)$. But I don't know how to resolve the question and what's utility of $\phi_1$ and $\phi_2$.
Thank's in advance to the help.
 A: Given the equation
$y'' + a(x)y = 0, \tag 1$
with Wronskian
$W[y_1, y_2] = y_1y_2' - y_1'y_2, \tag 2$
we have
$W' = (y_1y_2' - y_1'y_2)' = y_1'y_2' + y_1y_2'' - y_1''y_2 - y_1'y_2' = y_1y_2'' - y_1''y_2; \tag 3$
now, (1) implies
$y_i'' = -a(x)y_i, \; i = 1, 2; \tag 4$
substituting (4) into (3) we find
$W' = -a(x)y_1y_2 + a(x)y_1y_2 = 0; \tag 5$
it follows that $W$ is constant; thus if we initialize the $y_i$, $y_i'$, $i = 1, 2$ such that 
$W[y_1, y_2] = 1, \tag 6$
$W$ will remain fixed at $1$ over the entire range of $x$ for which the $y_i$ exist.  This may be accomplished by setting
$y_1(x_0) = 1 = y_2'(x_0), \tag 7$
and
$y_1'(x_0) = 0 = y_2(x_0); \tag 7$
then
$W[y_1, y_2](x_0) = 1, \tag 8$
whence
$\forall x, \; W[y_1, y_2](x) = 1, \tag 9$
provided the $y_i$ satisfy (1) with the stated conditions at $x_0$. 
Note Added in Edit, Friday 19 April 2019 8:41 AM PST:  The equation (1) is in fact a special case of the more general second order equation
$y'' + b(x)y' + a(x)y = 0; \tag{10}$
computing 
$W' = y_1y_2'' - y_1''y_2 \tag{11}$
using
$y_i'' = -b(x)y_i' - a(x)y_i, \; i = 1, 2, \tag{12}$
we find that instead of (5) we obtain
$W' = y_1(-b(x)y_2' - a(x)y_2) - (-b(x)y_1' - a(x)y_1)y_2 = -b(x)y_1y_2' + b(x)y_1'y_2$
$= -b(x)(y_1y_2' - y_1'y_2) = -b(x)W; \tag{13}$
the solution of this simple first-order, linear, homogeneous equation for $W(x)$ is well-known to take the form
$W(x) = \exp \left ( \displaystyle -\int_{x_0}^x b(s) \; ds \right ) W(x_0), \tag{14}$
from which we see that $W(x)$ will not in general be constant.  Indeed, if $W(x)$ is a constant, so that
$W(x) = W(x_0), \; \forall x \in I, \tag{15}$
then
$\exp \left ( \displaystyle -\int_{x_0}^x b(s) \; ds \right ) = 1, \; \forall x \in I, \tag{16}$
which implies that
$\displaystyle \int_{x_0}^x b(s) \; ds = 0, \forall x \in I; \tag{17}$
if $b(x)$ is assumed continuous we may differentiate this equation to obtain
$b(x) = 0, \forall x \in I, \tag{18}$
which shows that $W(x)$ is constant if and only if $b(x)$ vanishes.  End of Note.
A: You should have found out that the Wronskian is constant as the coefficient of the first derivative term is zero.
After that, it is just a matter of re-scaling one or both of the solutions to get the Wronski-determinant to have the value 1 at one and thus every point.

(Add) Interpreting the term "fundamental" in "system of fundamental solutions" more strictly, it means that at some point $x_0$ you have initial values
$$
\pmatrix{y_1(x_0)&y_2(x_0)\\y_1'(x_0)&y_2'(x_0)}
=\pmatrix{1&0\\0&1}
$$
so that no rescaling is necessary.
