Past exam of the differential equations 2 course of the University of Oxford / 2017 / 1 / a

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.

Differentiate the Wronskian using the product rule and substitute the common homogeneous linear differential equation:Multiplying this first-order equation by the integrating factor $\exp ({\int }_{x_0}^{x}p(t)dt)$ and integrating gives Abel's identity:The exponential factor is finite and nonzero for every $x$ in the given interval, because $p$ is continuous on the compact interval between $x_0$ and $x$. Hence $W(x_0)\ne 0$ implies $W(x)\ne 0$ everywhere in that interval. Using an integrating factor avoids dividing by $W$, which would assume the conclusion.