Suppose that $W(x_0) = 0$, and $y_1(x_0) = 0$. Show that $y_1(x)$=0 or $y_2(x) = \frac{y'_2(x_0)}{y'_1(x_0)} y_1(x)$ I know that
$W(x_0)=y_1(x_0)y'_2(x_0) -y'_1(x_0)y_2(x_0) = 0$ that implies 
$y'_1(x_0)y_2(x_0) = 0$,
If I supposed that $y_1$ and $y_2$ are solutions of $$y'' + p(x)y' + qy=0$$
I know that $y_1$ and $y_2$ are linearly depedent, then I know that $$y_1(x)y'_2(x) -y'_1(x)y_2(x)=0$$ that implies $$y_1(x)y'_2(x) =y'_1(x)y_2(x)$$ someone could give a hint?
 A: With the hypothesis you have stated this is obviously false. Take $x_0=0,y_1(x)=x,y_2(x)=x^{2}$ for a counterexample. I think $y_1$ and $y_2$ are suppose  to be solutions of some DE for such a result to be true.  
A: A rather challenging exercise, if I do say so myself, its resolution combining as it does techniques from existence and uniqueness, determinants and linear algebra, and specific, known forms of solution for first order equations.  To wit:
If $y_1(x)$ and $y_2(x)$ are solutions of the differential equation
$y''(x) + p(x)y'(x) + q(x)y(x) = 0, \tag 1$
linearly independent or not, we define their Wronskian matrix via
$W(y_1, y_2) = \begin{bmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{bmatrix}, \tag 2$
and the Wronskian determinant
$\det(W(y_1, y_2)) = \det \left ( \begin{bmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{bmatrix} \right ) = y_1(x)y_2'(x) - y_2(x)y_1'(x); \tag 3$
with the aid of (1), we may derive an expression for $(\det(W(y_1, y_2))'$, viz.
$(\det(W(y_1, y_2))' = (y_1(x)y_2'(x) - y_2(x)y_1'(x))'$
$= y_1'(x)y_2'(x) + y_1(x)y_2''(x) - y_2'(x)y_1'(x) - y_2(x)y_1''(x) = y_1(x)y_2''(x) - y_1''(x)y_2(x); \tag 4$
since $y_1(x)$ and $y_2(x)$ each satisfy (1), we find
$y_i''(x) = -p(x)y_i'(x) - q(x)y_i(x), \; i = 1, 2, \tag 5$
and substituting the equations (5) into (4),
$(\det(W(y_1, y_2))' = y_1(x)(-p(x)y_2'(x) - q(x)y_2(x)) - (-p(x)y_1'(x) - q(x)y_1(x))y_2(x) = p(x)(y_1'(x)y_2(x) - y_1(x)y_2'(x)) = -p(x) \det(W(y_1, y_2)); \tag 6$
the solution to this equation is, in general
$\det W(x) = \det(W(y_1(x), y_2(x)))$
$= \det W(x_0) e^{-\int_{x_0}^x p(s) \; ds} = \det W(x_0) \exp \left (-\displaystyle \int_{x_0}^x p(s) \; ds \right ), \tag 7$
as the reader may easily verify, and since
$\exp \left (-\displaystyle \int_{x_0}^x p(s) \; ds \right ) \ne 0, \forall x, \tag 8$
it follows that
$\det W(x) = 0 \Longleftrightarrow \det W(x_0) = 0 \tag 9$
for some $x_0$.
Now granted that
$\det W(x_0) = 0, \tag{10}$
we have seen that
$\det W(x) = 0, \forall x; \tag{11}$
if we further hypothesize that
$y_1(x_0) = 0, \tag{12}$
then if
$y_1(x) \ne 0, \forall x, \tag{13}$
we must have
$y_1'(x_0) \ne 0; \tag{14}$
otherwise, 
$y_1(x) = 0, \forall x \tag{15}$
follows from uniqueness of solutions of (1);  furthermore, by virtue of (3) and (11)
$y_1(x)y_2'(x) - y_2(x)y_1'(x) = 0, \forall x; \tag{16}$
from this equation we infer that the rows and columns of the matrix $W(y_1, y_2)$ in (2) are linearly dependent for all $x$; in particular we have for $x_0$:
$y_1(x_0)y_2'(x_0) - y_2(x_0)y_1'(x_0) = 0, \tag{17}$
whence (14) allows us to write
$y_2(x_0) = \dfrac{y_2'(x_0)}{y_1'(x_0)}y_1(x_0), \tag{18}$
or
$y_2(x_0) - \dfrac{y_2'(x_0)}{y_1'(x_0)}y_1(x_0) = 0; \tag{19}$
since $y_1(x)$ and $y_2(x)$ satisfy (1), it follows from linearity that the function
$y_2(x) - \dfrac{y_2'(x_0)}{y_1'(x_0)}y_1(x) \tag{20}$
is also a solution to (1); we also have
$\left ( y_2(x) - \dfrac{y_2'(x_0)}{y_1'(x_0)}y_1(x) \right )' = y_2'(x) - \dfrac{y_2'(x_0)}{y_1'(x_0)}y_1'(x) = 0, \; x = x_0; \tag{21}$
we now once again invoke uniqueness of solutions of the linear equation (1) to conclude that the solution (20) vanishes for all $x$; that is
$y_2(x) = \dfrac{y_2'(x_0)}{y_1'(x_0)}y_1(x) = 0, \; \forall x, \tag{22}$
as was to be proved.
