Different integral equations on solution of Sturm-Liouville equation Consider a differential equation
$$
   u''(x) + \lambda^2 u(x) = q(x)u(x), \;\;\; x>0, \; \Im\lambda \geqslant 0, \; \lambda \neq 0
$$
with boundary condition $u(0)=0$. Here potential $q(x)$ is continuous and $q(x) \in L_1\left([0,\infty],1+x\right)$. I solved a problem
$$
   \left\{ \begin{array}{c}
   v''(x) + \lambda^2 v(x) = 0, \\
   v(0) = 0, \; v'(0) = 1. \end{array} \right.
$$
The solution is $v(x) = \frac{\sin(\lambda x)}{\lambda}$. Then for $L = \frac{d^2}{dx^2}+\lambda^2$ we have $L(v(x)\chi(x)) = \delta(x)$. Then
$$
   u(x) = (v\chi)*(qu)(x)=\int\limits_{0}^{x}\frac{\sin \lambda(x-t)}{\lambda}q(t)u(t)dt
$$
satisfies a differential equation $Lu(x) = q(x)u(x)$. But I was told that there is an another integral equation on $u$ (maybe, different from the above $u$):
$$
   u(x) = e^{i \lambda x}+\int\limits_{x}^{\infty} \frac{\sin \lambda(t-x)}{\lambda} q(t)u(t)dt.
$$
Please, help me to obtain it.
 A: I found a way of receiving such solutions. Let $u_0(x)$ be a solution of homogenous system $Lu_0 = 0$. We can choose such $u_0(x)$ with appropriate asymptotics at zero or at infinity. Next, if we want to obtain a solution of system $Lu = qu$ with asymptotics of $u_0(x)$ at zero (or at infinity) we can define $u(x)$ ($\tilde{u}(x)$) by the integral equation
$$
    u(x) = u_0(x)+\int\limits_{0}^{x} \frac{\sin \lambda(x-t)}{\lambda} q(t) u(t) \, dt \\
  \left(  \tilde{u}(x) = u_0(x)+\int\limits_{x}^{\infty} \frac{\sin \lambda(t-x)}{\lambda} q(t) \tilde{u}(t) \, dt \right)
$$
Then 
$$
Lu = Lu_0 + L( v \chi* qu) = 0 + L(v\chi)*qu = 0 + \delta * qu = qu \\
\left( L\tilde{u} = Lu_0 + L(v(-x)\chi(-x)*q\tilde{u}) = 0 + \delta*q\tilde{u} = q\tilde{u} \right)
$$
and $u$ ($\tilde{u})$ has the same asymptotics at zero (infinity) as $u_0$. Particularly, if $u_0(x) = 0$ then
$$
   u(x) = \int\limits_{0}^{x} \frac{\sin \lambda(x-t)}{\lambda} q(t) u(t) \, dt
$$
solves $Lu = qu$ with initial condition $u(0)=0$ and if $u_0(x) = e^{i \lambda x}$ then
$$
  \tilde{u}(x) = e^{i \lambda x} + \int\limits_{x}^{\infty} \frac{ \sin \lambda(t-x)}{\lambda} q(t) \tilde{u}(t) \, dt
$$
solves $L\tilde{u} = q\tilde{u}$ with asympotics $\tilde{u} \sim e^{i \lambda x}$ when $x \to + \infty$.
