Solve integral equation by converting to differential equation

Question

One has the following integral equation:

$$y(x)=\frac{1}{1+x^2}+\int_{0}^{x}\sin(x-t)y(t)dt$$

How can I solve this integral equation by converting it to a differential equation?

Answer 1

Let $g(x)=\dfrac{1}{1+x^2}$. Then, we have

$$y(x)=g(x)+\int_0^x \sin(x-t)y(t)dt \tag 1$$

Differentiating both sides of $(1)$ yields

$$y'(x)=g'(x)+\int_0^x \cos(x-t)y(t)dt \tag 2$$

Differentiating both sides of $(2)$ reveals

$$\begin{align} y''(x)&=g''(x)+y(x)-\int_0^x \sin(x-t)y(t)dt \tag 3\\\\ &=g''(x)+g(x) \tag 4 \end{align}$$

where we used $(1)$ $(3)$ to arrive at $(4)$. Thus, upon integrating both sides of $(4)$ we find

$$\bbox[5px,border:2px solid #C0A000]{y(x)=\frac{1}{1+x^2}+x\arctan(x)-\frac12 \log(x^2+1)} \tag 5$$

where we used $y(0)=1$ and $y'(0)=0$ as provided by $(1)$ and $(2)$ to obtain $(5)$.

Answer 2

if we assign the second part on the right side as $I$ we can quickly apply Differentiation of integral to arrive at $I^{''} = -I $

observing that $I = y(x)-\frac{1}{1+x^2}$ we can reduce the final differential equation to $y^{''}+y - 1/(1+x^2) -d^2(1/(1+x^2))/dx^2=0 $

now substitute $z = y - 1/(1+x^2)$ . hence $z^{''} + z = 0 $