Question

Let $f:[0,\infty )\to R$ be a continuous function such that $f\left(x\right)=1-2x+\underset{0}{\overset{x}{\int }}{e}^{x-t}f\left(t\right)dt$ for all $x\in [0,\infty ).$ Then the value of $f(-5)$ is

Answer 1

$f\left(x\right)=1-2x+e^x\underset{0}{\overset{x}{\int }}{e}^{-t}f\left(t\right)dt\cdots \left(i\right)$
$\Rightarrow f^{\prime }\left(x\right)=-2+e^x\underset{0}{\overset{x}{\int }}{e}^{-t}f\left(t\right)dt+e^x{e}^{-x}f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=-2+f\left(x\right)-1+2x+f\left(x\right)\left[\text{From}\right(i\left)\right]$
$\Rightarrow f^{\prime }\left(x\right)-2f\left(x\right)=2x-3$
which is a linear differential equation.

$\text{IF}=e^{-2x}$
$\therefore f\left(x\right)\cdot e^{-2x}=\int (2x-3)e^{-2x}dx$
$\Rightarrow f\left(x\right)\cdot e^{-2x}=\frac{(2x-3)e^{-2x}}{-2}-\frac{e^{-2x}}{2}+C$

$\text{From}\left(i\right),f\left(0\right)=1$
$\Rightarrow f\left(x\right)=1-x$
$\Rightarrow f(-5)=6$