Differential equations, integral equations

Question

Is there an analytical way of proving that if $\phi$ is a solution to \begin{equation} y(t)=e^{it}+a\int_{t}^{\infty}\sin (t-s)y(s)s^{-2}ds, \end{equation} then $\phi$ would be a solution to the differential equation \begin{equation} y''+(1+at^{-2})y=0\quad ? \end{equation}

Answer 1

Here is how you advance. We have

$$ \begin{equation} y(t)=e^{it}+a\int_{t}^{\infty}\sin (t-s)y(s)s^{-2}ds, \end{equation}. $$

Now, differentiating the above equation twice with respect to $t$ yields

$$y''= -e^{it} - a \int _{t}^{\infty }\!{\frac {\sin \left( t-s \right) y \left( s \right) }{{s}^{2}}}{ds}-a{\frac {y \left( t \right) }{{t}^{2} }} .$$

Adding the two equations and simplifying gives you the desired result. See here.

Note: For differentiating under the integral sign, we need the Leibniz rule

$$ \frac{d}{dx} \left (\int_{a(x)}^{b(x)}f(x,t)\,dt \right) = f(x,b(x))\,b'(x) - f(x,a(x))\,a'(x) + \int_{a(x)}^{b(x)} f_x(x,t)\; dt. $$