The solution of differential equation `(dy)/(dx) + xy = xy^(2)` is

Options

A. ` -1/y = 1+Ce^(x^(2)//2`
B. `1/y= 1+Ce^(x^(2)//2)`
C. ` 1/y = 1- Ce^(x^(2))`
D. None of the above

Answer

b

Explanation

Given that , `(dy)/(dx) + xy = xy^(2)" "` …(i)
This is type of Bernaulli's equation . So divide Eq. (1) by `y^(2)`
` :. y^(-2) (dy)/(dx) +1/y * x = x `
Put `1/y = v`
` rArr -1/(y^(2)) (dy)/(dx) = (dy)/(dx)`
` :. (dv)/(dx) - xv = - x`
On comparing above equation with `(dv)/(dx) +Pv = Q` where P
and Q are functions of x only .
` P = -x , Q = - x`
` IF = e^(int -xdx) = e^(-x^(2)//2)`
` :. " The solution is " v xx IF = int (Q xx IF ) dx + C`
` ve^(-x^(2)/2) = int - x e ^(-x^(2)/2) dx = e^(-x^(2)/2) +C`
` rArr 1/y = 1 + C e ^(2//2)`