I spotted the following lemma in a paper titled "On Approximation by Solutions of Ordinary Linear Differential Equations" in J. SIAM Numer. Anal. Vol. 3, No. 2, 1966. It's provable through a straightforward application of Rolle's theorem after discovering an appropriate integrating factor. Let's dig in.
Lemma. Let \(f(x)\) be differentiable in the compact interval \([a,b]\) and suppose \(f(a) = f(b) = 0\). Then for every \(\lambda (x)\in C[a,b]\) there exists \(c=c(\lambda )\), \(a<c<b\), such that \[ f^\prime (c) + \lambda (c)f(c) = 0 \]
Proof. Notice that \[ f^\prime (x) + \lambda (x)f(x) = 0 \] is a first-order linear differential equation of the form \[ y^\prime + P(x)y = Q(x). \]
Differential equations of this sort are amenable to solution by discovering an integrating factor \(M(x)\) so that \[ f^\prime (x)M(x) + \lambda (x)f(x)M(x) = \frac {dF}{dx} \] for some function \(F\).
Let \(\lambda (x)\) be given as above. Define
\[ F(x) = f(x)\exp \left (\int _a^x\lambda (t)dt\right ) \]
and observe that \(F(a) = F(b) = 0\) since \(f(a) = f(b) = 0\). \(F(x)\) is the product of two continuous and differentiable
functions. Thus \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\).
This function, then, satisfies the conditions for Rolle’s theorem. Then it must be that there exists some \(c \in (a,b)\) with \[ F^\prime (c) = 0 = f^\prime (c)\exp \left (\Lambda (c)\right ) + f(c)\lambda (c)\exp \left (\Lambda (c)\right ) \] where \(\Lambda (x) = \int _a^x\lambda (t)dt\) is an antiderivative of \(\lambda \). Since \(\exp (z)\) is non-zero for all \(z\), we must have \[ f^\prime (c) + f(c)\lambda (c) = 0 \] which is the desired result. □