Consider a non-zero, real-valued polynomial function $p(x)=a+bx+cx^2$ of degree at most 2. Let $y(x)$ be any solution of integral equation $$ y (x)=p(x) + \int_{0}^{x}y(t)\sin (x-t)dt . $$ Then which of the following is true .
- (1) $y (x)$ is polynomial function of degree less than or equal to 2 .
- (2) $y (x)$ is polynomial function of degree less than or equal to 4.
(3) if $b$ is non zero and $a+2c$ is zero then $y'(0)=0$.
(4) if b is non zero and $a+2c$ is zero then $y''(0)=0$.
Using Laplace transformation, i get solution $$ y (x)=a +bx +(c/2 + a)x^2 +(a/6)x^3 +(c/12)x^4, $$ so option( $2$) is correct. If $b$ is non zero then $y'(0)$ is non zero, but $y''(0) =0$ , so reject (3). So(2) and (4)are correct am I right? But I am also confused about (1) may be there is some polynomial $p (x)$ for which option (1) is true . But in Question we have to find which is " Necessary" true . So option (1) is wrong . Am I right ?
The integral equation $$ y(x)=p(x)+\int_0^xy(t)\sin(x-t)dt $$ has derivatives $$ y'(x)=p'(x)+\int_0^xy(t)\cos(x-t)dt\\ y''(x)=p''(x)+y(x)-\int_0^xy(t)\sin(x-t)dt $$ It is thus equivalent to the second order ODE $$ y''(x)=p(x)+p''(x) $$ with initial conditions $y(0)=p(0)=a$ and $y'(0)=p'(0)=b$. Further, $y''(0)=p''(0)+y(0)=2c+a$.
This should be sufficient to decide all claims, and indeed your decisions are right. (1) would have to be true for all $p$, which is not the case.