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I was looking for the solution of the following problem.

Prove that if $\phi$ is a solution of the integral equation

$$y(t) = e^{it} + \alpha \int\limits_{t}^\infty \sin(t-\xi)\frac{y(\xi)}{\xi^2}d\xi,$$

then $\phi$ satisfies the differential equation

$$y'' + (1+\frac{\alpha}{t^2})y=0$$

Do I need to solve the differential equation to get the integral equation or I have to solve the integral equation to get the differential equation.

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  • $\begingroup$ Don't use pictures, it makes the question unsearchable for future users. Write the problem in LaTeX next time. $\endgroup$
    – TSF
    Feb 20 '18 at 14:41
  • $\begingroup$ Sorry. Actually I am new here. So I face difficulty to write the question. $\endgroup$
    – user496876
    Feb 20 '18 at 14:52
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    $\begingroup$ Don't worry I transcribed it for you, just a reminder for next time. $\endgroup$
    – TSF
    Feb 20 '18 at 14:56
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    $\begingroup$ Leibniz integral rule is what you need here: $\frac{d}{dt}\int_t^\infty f(t,\xi){\rm d}\xi = -f(t,t) + \int_t^\infty \frac{\partial f(t,\xi)}{\partial t}{\rm d}\xi$. Use this (twice) to compute $y''$ and compare to $y$. $\endgroup$
    – Winther
    Feb 20 '18 at 14:57
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Take the derivative of $y$ twice using the equation,

$$y(t) = e^{it} + \alpha\int\limits_{t}^\infty \sin(t-\xi)\frac{y(\xi)}{\xi^2}d\xi$$

Then plug what you get into

$$y''+(1+\frac{\alpha}{t^2})y$$

and simplify to get $0$ and thus show the two are equivalent.

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  • $\begingroup$ I tried it. When I take the derivative of second part of y. Will the integration will cancel the derivative and I have to apply the limits from t to infinity. $\endgroup$
    – user496876
    Feb 20 '18 at 14:55
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    $\begingroup$ No because you are taking the derivative with respect to $t$, see the following: en.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$
    – TSF
    Feb 20 '18 at 14:57
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Using $\frac{d}{dt}\int_t^\infty f(t,\xi){\rm d}\xi = -f(t,t) + \int_t^\infty \frac{\partial f(t,\xi)}{\partial t}{\rm d}\xi$, as was suggested many times above, you have $$\frac{d}{dt}y=\frac{d}{dt}\Big(e^{it} + \alpha \int\limits_{t}^\infty \sin(t-\xi)\frac{y(\xi)}{\xi^2}d\xi\Big)=ie^{it}+\alpha\int\limits_{t}^\infty \cos(t-\xi)\frac{y(\xi)}{\xi^2}d\xi$$ and $$\frac{d^{2}}{dt^{2}}y=\frac{d}{dt}\Big(ie^{it}+\alpha\int\limits_{t}^\infty \cos(t-\xi)\frac{y(\xi)}{\xi^2}d\xi\Big)=-e^{it}-\alpha\frac{y(t)}{t^{2}}-\alpha\int\limits_{t}^\infty \sin(t-\xi)\frac{y(\xi)}{\xi^2}d\xi$$ And using $$e^{it}+\alpha\int\limits_{t}^\infty \sin(t-\xi)\frac{y(\xi)}{\xi^2}d\xi=y$$ You get $$\frac{d^{2}}{dt^{2}}y=-y-\alpha\frac{y}{t^{2}}$$

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