Abel´s formula solutions of the DE If  $y_{1}$  and $y_{2}$ are linearly independent solutions of the second order DE
${y}''+a_{1}(x){y}'+a_{0}(x)y=0$
on some interval where $a_{1}(x)$ and $a_{0}(x)$ are continuous, show that the Wronskiant satisfies, for some constant $C$.
$W(y_{1},y_{2})(x)=C exp\left [ -\int a_{1}(x)dx \right ]$
Hint: Show that  $\frac{\mathrm{dW} }{\mathrm{d} x}+a_{1}(x)W=0$, and solve this  DE for $W$
Any kind of help will be appreciated. Thank you.
 A: We have
$W(y_1, y_2) = \det \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} = y_1y_2' - y_2y_1'; \tag 1$
then
$W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2 y_1'' = y_1y_2'' - y_2y_1'' ; \tag 2$
from the given differential equation
$y'' + a_1(x) y' + a_0(x)y = 0, \tag 3$
we have
$y_i'' =  -a_1(x) y_i' - a_0(x)y_i, \; i = 1, 2; \tag 4$
substituting these into (2) yields
$W' = y_1(-a_1(x) y_2' - a_0(x)y_2) - y_2(-a_1(x) y_1' - a_0(x)y_1) = -a_1(x)y_1y_2' -a_0(x)y_1y_2 + a_1(x) y_2y_1' + a_0y_1y_2 = -a_1(x) (y_1y_2' - y_1'y_2) = -a_1(x)W, \tag 5$
that is
$W' + a_1(x)W = 0, \tag 6$
the solution of which is well-known to be
$W(x) = W(x_0)\exp \left ( -\displaystyle \int_{x_0}^x a_1(s) \; ds \right ); \tag 7$
setting
$C= W(x_0) = W(y_1(x_0), y_2(x_0)) \tag 8$
yields
$W(y_1(x), y_2(x)) = W(x) = W(x_0)\exp \left ( -\displaystyle \int_{x_0}^x a_1(s) \; ds \right ), \tag 9$
the requisite result.  $OE\Delta$.
A: Since 
$W = \begin{vmatrix}
y_1 & y_2 \\ 
y_1' & y_2' \\ 
\end{vmatrix} = y_1 y_2' - y_2 y_1'$
Hence
$\frac{dW}{dx} + a_1(x) W = y_1' y_2' + y_1 y_2'' - y_2' y_1' - y_2 y_1'' + a_1(x)(y_1 y_2' - y_2 y_1') = y_1 (y_2'' + a_1(x) y_2' + a_0 y_2) - y_2 (y_1'' + a_1(x) y_1' + a_0 y_1) = 0$
