Question

Let $f$ be a derivative function satisfying
$f\left(x\right)=x^2+{\int }_0^x{e}^{-t}f(x-t)dt$ then degree of polynomial function $f\left(x^2\right)$ is

Answer 1

Given:
$f\left(x\right)=x^2+{\int }_0^x{e}^{-t}f(x-t)dt$
To find degree of $f\left(x^2\right)$
solution;
$f\left(0\right)=0+{\int }_0^0{e}^{-t}f(-t)dt=0f\left(t\right)=t^2+{\int }_0^t{e}^{(-t)}f(t-t)dt=t^2+{\int }_0^t{e}^{(-t)}f\left(0\right)dt=t^2+0[\because f/0=0]=f\left(t\right)=t^2\Rightarrow f\left(x\right)=x^2\therefore t\left(x^2\right)=x^4$
Hence, degree of the polynomial $f\left(x^2\right)=4$