For the equation: $${-y'' + q(x) y = k^2 y}$$
whith the initial conditions:
$$y(0;k) = 0 $$
$$y'(0;k)= 1 $$
Its equivalent integral equation is:
$${y(x; k)} = {\frac {sin(k x)}{k} + \int^x_0 {\frac{sin(k(x - t))}{k}}q(t) y(t;k) dt}$$
I'm trying to demostrate that for this equation, the estimation:
$$|| y(x;k) || \leqslant K \frac{x e^{\tau x}}{1 + |k| x}$$
holds up for $k = \sigma + i \tau $ ; $\sigma, \tau \in \mathbb{R} $ ; $x \geqslant 0$, where $ K $ is a constant
From the succesive approximations method, I get that for the equation: $${\phi(x; \lambda)} = {f(x) + \lambda \int_G {k(x,y) \phi(y;\lambda) dy}}$$ $$||\phi(x;\lambda)|| \leqslant \frac{||{f(x)}||_c}{1 - M |\lambda| V} $$
where:
$$M = {max_{x \in \bar G} |k(x,y)}|$$
$$ V = \int_G dy $$
I'm having trouble with calculating $M$. Is it possible to make the proof without knowing $q(x)$ explicitly?