Given $y_k=J_m(\sqrt{\lambda_k}x)$ and let $y(x,\lambda)=J_m(\sqrt{\lambda}x)$.
I can't seem to compute this integration and show $\int^1_0({{\dfrac{d}{dx}(xy'_k)y-\dfrac{d}{dx}(xy')y_k}} )dx={\sqrt{\lambda_k}J'_m(\sqrt{\lambda_k})J_m(\sqrt{\lambda})}$
Integration by parts leads to:
$$\int_{0}^{1}\frac{d}{dx}(x y_k') y \, dx = \left. x y_k' y\right|_{0}^{1}-\int_{0}^{1}x y_k' y'\,dx $$
$$\int_{0}^{1}\frac{d}{dx}(x y') y_k \, dx = \left. x y' y_k\right|_{0}^{1}-\int_{0}^{1}x y_k' y'\,dx $$ hence:
$$\int_{0}^{1}\left(\frac{d}{dx}(x y_k') y-\frac{d}{dx}(x y') y_k\right)\,dx = \left.\left(x y y_k'- x y_k y'\right)\right|_{0}^{1} = \left(yy_k'-y_k y'\right)(1).$$