If solution of the differential equation $\frac{dy}{dx}=y\left(x\right)+{\int }_0^{1}y\left(x\right)dx$  and $y\left(0\right)=1$  is $y\left(x\right)$ , then the value of $\left[y\left(1\right)-7\right]$  (where $\left[.\right]$  denotes the greatest integer function) is

The given differential equation can be written as  $y\text{'}\left(x\right)=y\left(x\right)+{\int }_0^{1}y\left(x\right)dx$
Differentiating both sides w.r.t.  x  then 
  $y"\left(x\right)=y\text{'}\left(x\right)$
Or  $\frac{y"\left(x\right)}{y\text{'}\left(x\right)}=1$
Or  $\int \frac{y"\left(x\right)}{y\text{'}\left(x\right)}dx=\int dx$
Or ln  $y\text{'}\left(x\right)=x+\ln c$
Or  $\ln \left(\frac{y\text{'}\left(x\right)}{c}\right)=x$
Or  $y\text{'}\left(x\right)=c{e}^x$
And  $y\left(x\right)=c{e}^x+d$
 $\Rightarrow y\left(0\right)=c+d=1\left(\because y\left(0\right)=1\right)$
 $\therefore y\left(x\right)=c{e}^x+1-c$
From (i), (ii) and (iii), we get 
  $c{e}^x=c{e}^x+1-c+{\int }_0^1\left(c{e}^x+1-c\right)dx$
 $\Rightarrow c-1={\left[c{e}^x+\left(1-c\right)x\right]}_0^1$
 $\Rightarrow c-1=ce+1-c-c$
 $\therefore c=\frac{2}{3-e}$
From (iii), $y\left(1\right)=ce+1-c=\frac{e+1}{3-e}$                           [From (iv)]

$=\frac{2718+1}{3-2718}=\frac{3.718}{0.282}=13.18$
 $\left[y\left(1\right)-7\right]=\left[6.18\right]=6$