Consider the differential equation, `y^(2)dx+(x-1/y)dy=0.` ltbegt If value of y is 1 when x = 1, then the value of x for which
y = 2, is

Options

A. `(3)/(2)-sqrte`
B. `(1)/(2)+(1)/(sqrte)`
C. `(5)/(2)+(1)/(sqrte)`
D. `(3)/(2)-(1)/(sqrte)`

Answer

D

Explanation

`y^(2)dx+(x-(1)/(y))dy=0`
`y^(2)(dx)/(dy)+x-(1)/(y)=0," "(dx)/(dy)+(x)/(y^(2))=(1)/(y^(2))`
IF `=e^(int(1)/(y^(2))dy)=e^(-(1)/(y))`
`therefore" Solution is x.lF "=int(1)/(y^(3)).e^(-(1)/(y))=int-t.e^(t)dt" "("Put "-(1)/(y)=t)`
And apply by parts and replacing t we get
`xe^(-(1)/(y))=+(1)/(y)e^(-(1)/(y))+e^(-(1)/(y))+e" "x=((1)/(y)+1)+ce^((1)/(y))`
`"Given "x=1" "y=1" "rArr" "C=(-1)/(c)`
`"Now at "y=2" "x=(3)/(2)+(1)/(e)(e^((1)/(2)))=(3)/(2)-(1)/(sqrte)`