Solution to the integral equation with constant parameter

Question

I wanted to solve this integral equation, I'll be grateful for explanation with step by step solution.

$y(x)=1+α\int_{0}^xdp sin(x-p)y(p)$

Answer 1

Hint for first steps.

\begin{align*}y(x)&= 1+\alpha \int_{0}^x \sin(x-p)y(p) dp \\ &= 1+\alpha \int_{0}^x (\sin x \cos p - \cos x \sin p)y(p) dp \\ &= 1+\alpha \sin x\int_{0}^x (\cos p) y(p) dp - \alpha \cos x \int_0^x (\sin p) y(p) dp \\ & = 1 + \alpha \sin x C(x) - \alpha \cos x S(x) \end{align*} where $C(x) = \int_{0}^x (\cos p) y(p) dp $ and $S(x) = \int_{0}^x (\sin p) y(p) dp $.

Take derivative $d/dx$ to have \begin{align*}y'(x) &= \alpha \cos x C(x) + \alpha (\sin x \cos x) y(x)+ \alpha \sin x S(x) - \alpha (\cos x \sin x) y(x) \\ & = \alpha \left( \cos x C(x) + \sin x S(x) \right) \end{align*}

Take derivative $d/dx$ once more to have \begin{align*} y''(x) & = \alpha \left( -\sin x C(x) + (\cos x \cos x) y(x) + \cos x S(x) + (\sin x \sin x) y(x)\right) \\ & = \alpha y(x) - \alpha \sin x C(x) + \alpha \cos x S(x)\\ & = \alpha y(x) + 1 -y(x) = 1 - (1- \alpha) y(x) \end{align*}

Can you proceed from here?

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Edit. I will proceed.

This is linear differential equation with constant coefficient and is not homogeneous. There is homogeneous solution $y_h$ for $y'' = (\alpha-1)y$, and a particular solution $y_p$ satisfying $y'' = (\alpha-1) y +1$, which are summed up to form a solution $y= y_h + y_p$.

By assuming $\alpha \ne 1$, it is easy to see that $y_h = A e^{\sqrt{\alpha-1}x} + B e^{-\sqrt{\alpha-1}x}$ for constants $A, B$. (Is it easy for you?)

For $y_p$, there is one trivial solution $y_p(x) = \frac{1}{1-\alpha}$. So the solution is of the form $$y(x) = A e^{\sqrt{\alpha-1}x} + B e^{-\sqrt{\alpha-1}x} +\frac{1}{1-\alpha}.$$

If $\alpha <1$, one might feel more comfortable with $$y(x) = A_0 \cos{\sqrt{1-\alpha}x} + B_0 \sin{\sqrt{1-\alpha}x} +\frac{1}{1-\alpha}.$$

Initial conditions are implicitly given; one can compute $y(0)$ and $y'(0)$. I will left this part for you.

Finally, if $\alpha = 1$, we have $y''(x) = 1$ and solution is simply $y = \frac{x^2}{2} + Ax + B$.