$f:[0,\infty )\to R$ is 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 which of following statement is/are TRUE?

$f\left(x\right)=1-2x+\underset{0}{\overset{x}{\int }}{e}^{x-t}f\left(t\right)dt ,x\in [0,\infty )$

$$\begin{gathered} f\left(x\right)=1-2x+\underset{0}{\overset{x}{\int }}{e}^x{e}^{-t}f\left(t\right)dt, \\ f\left(x\right)=(1-2x)+e^x\underset{0}{\overset{x}{\int }}{e}^{-t}f\left(t\right)dt \end{gathered}$$

Differentiate w.r.t x
 $f\text{'}\left(x\right)=-2+e^x\underset{0}{\overset{x}{\int }}{e}^{-t}f\left(t\right)dt+e^x\left(e^{-x}f\left(x\right)\right)$

$f\text{'}\left(x\right)=2f\left(x\right)+2x-3$

$$\begin{gathered} \frac{dy}{dx}=2y+(2x-3), \\ \frac{dy}{dx}-2y=(2x-3) \end{gathered}$$

 $I.F=e^{\int -2dx}=e^{-2x}$  
Solution given by y. $e^{-2x}=\int (2x-3)e^{-2x}dx$

$y{e}^{-2x}=(2x-3)\frac{e^{-2x}}{-2}+\int e^{-2x}dx$

$y{e}^{-2x}=\frac{-1}{2}(2x-3)e^{-2x}+\frac{e^{-2x}}{-2}+C$

$y=-\frac{1}{2}(2x-3)-\frac{1}{2}+C{e}^{2x}$

For $x\in [0,\infty )$ at x = 0, y = 1

$y=-\frac{1}{2}(-3)-\frac{1}{2}+C$

$$\begin{gathered} 1=\frac{3}{2}-\frac{1}{2}+C \\ \Rightarrow 1=1+C\to \boxed{C=0} \end{gathered}$$

$$\begin{gathered} y=-\frac{1}{2}(2x-3)-\frac{1}{2}\Rightarrow -2y=2x-3+1 \\ \Rightarrow -2y=2x-2 \\ \Rightarrow -y=x-1 \\ \Rightarrow \boxed{y=-x+1} \end{gathered}$$