If y, and y, are linearly dependent solutions of the differential equation a(x)y" + a,(x)y' + a,(x)y = y(x), show that W(y,, y,) = 0.

Two functions $y_1(x)$ and $y_2(x)$ are linearly dependent on an interval $I$ if there exist constants $c_1$ and $c_2$, not both zero, such that for all $x \in I$:

$$c_1 y_1(x) + c_2 y_2(x) = 0$$

Since $c_1$ and $c_2$ are not both zero, we can assume without loss of generality that $c_1 \neq 0$. (If $c_1=0$, then $c_2 \neq 0$, and we could express $y_2$ in terms of $y_1$ instead).
From the linear dependence relation, we can write:

$$c_1 y_1(x) = -c_2 y_2(x)$$ $$y_1(x) = -\frac{c_2}{c_1} y_2(x)$$

Let $k = -\frac{c_2}{c_1}$. Then $y_1(x) = k y_2(x)$.
Differentiating both sides with respect to $x$, we get:

$$y_1'(x) = k y_2'(x)$$

The Wronskian of $y_1(x)$ and $y_2(x)$ is defined as the determinant:

$$W(y_1, y_2)(x) = \begin{vmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{vmatrix}$$

Substitute $y_1(x) = k y_2(x)$ and $y_1'(x) = k y_2'(x)$ into the determinant:

$$W(y_1, y_2)(x) = \begin{vmatrix} k y_2(x) & y_2(x) \\ k y_2'(x) & y_2'(x) \end{vmatrix}$$

Calculate the determinant:

$$W(y_1, y_2)(x) = (k y_2(x))(y_2'(x)) - (y_2(x))(k y_2'(x))$$ $$W(y_1, y_2)(x) = k y_2(x)y_2'(x) - k y_2(x)y_2'(x)$$ $$W(y_1, y_2)(x) = 0$$

Thus, if $y_1$ and $y_2$ are linearly dependent, their Wronskian is zero.

The final answer is $W(y_1, y_2) = 0$.