If the solution of the differential equation `(dy)/(dx)-y=1-e^(-x)` and `y(0)=y_0` has a finite value, when `xvecoo,` then the value of `|2/(y_0)|` is__

Answer

4

Explanation

`(dy)/(dx) -y=1-e^(-x)`
`P=-1,Q=1-e^(-x)`
I.F`. E^(intPdx)=e^(int-1dx)=e^(-x)`
`therefore y.e^(-x)=inte^(-x)(1-e^(-x))dx+C`
`=e^(-x)+1/2e^(-2x)+C`
When, `x=0, y=y_(0)+1/2`
So, `C=y_(0)+1/2`
`y=-1+1/2e^(-x) +(y_(0)+1//2)e^(x)`
When `x to infty y` takes finite value.
So, `y_(0)+1//2=0`
or `y_(0)=-1//2`