 Solution

Extend the given functions continuously to zero by $w_1(0)=w_2(0)=0$. This is forced by their stated differentiability. Their derivatives at zero are both zero, since the corresponding difference quotients tend to zero.

For $x<0$, $w_2=w_2^{\prime }=0$, and for $x>0$, $w_1=w_1^{\prime }=0$. At zero all four quantities vanish. Thus the Wronskian isNevertheless, if the constant linear combination $A{w}_1+B{w}_2$ vanishes identically, evaluation at $x=-1$ gives $-A=0$, and evaluation at $x=1$ gives $B=0$. Therefore the two functions are linearly independent.

This illustrates why a vanishing Wronskian does not imply linear dependence for arbitrary differentiable functions. Here both functions are continuously differentiable, but neither has a second derivative at zero. The common regular differential equation hypothesis in the next part is absent.