# Why does $W[y_1, y_2] = 0$ (wronskian) imply the following?

Consider a second order, linear, homogeneous differential equation of the form $$y''(x) + p(x) y'(x) + q(x) y(x) = 0$$ where $p(x)$ and $q(x)$ are continuous on some interval $I$.

Now let $x_0 \in I$. Then, if $y_1, y_2$ are solutions to this DE, and $W[y_1, y_2] = 0$ (if the wronskian of these functions is equal to zero), then this imples that there exists some non-zero values $c_1, c_2$ such that $$c_1 y_1(x_0) + c_2 y_2(x_0) = 0 \hspace{5 mm} \text{and} \hspace{10 mm} c_1 y_1'(x_0) + c_2 y_2'(x_0) = 0$$

Why does $W[y_1, y_2] = 0$ imply the above result?

## 2 Answers

Hint: The Wronskian satisfies : $W(t)=W(s)\exp(−\int_s^t p(x)ds$, for any $s,t\in I.$ Use this to show that $W[y_1, y_2](t)=0$ and proceed from there.

For differential equation $$y''(x) + p(x) y'(x) + q(x) y(x)=0\tag{1}$$ where $p(x)$ and $q(x)$ are continuous on some interval $I$.

Your question is equivalent with this

If $y_1(x_0)$ and $y_2(x_0)$ are independents solution of $(1)$ in $I$ then $W[y_1, y_2]\neq0$ , this is for $W(y_1,y_2)=‎\left|\begin{array}{rr}y_1&y_2\\y_1^\prime&y_2^\prime\end{array}\right|$ so ‎$$W^\prime=y_1y_2^{\prime\prime}-y_1^{\prime\prime}y_2=y_1(-py_2^\prime-qy_2)-y_2(-py_1^\prime-qy_1)=-pW$$ or $$\frac{dW}{dx}+pW=0$$thus $W=Ce^{-\int p(x)dx}\neq0$