If `dy/dx=e^(-2y) and y=0` when `x=5,` then the value of x for `y=3` is

Explanation

Given that, `(dy)/(dx)=e^(-2 y)Rightarrow (dy)/(e^(-2y))=dx`
`Rightarrow inte^(2y)=intdx Rightarrow (e^(2y))/(2)=x+C ....(i)`
When x=5 and y=0, then substituting these values in Eq. (i), we get
`(e^(0))/(2)=5+C`
`Rightarrow (1)/(2)=5+C Rightarrow (1)/(2)-5=(9)/(2)`
`"Eq. = (i) becomes e"^(2y)=2x-9`
When, y=3, then `e^(6)=2x-9 Rightarrow 2x=e^(6)+9`
`x=((e^(6))+9)/(2)`