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

Options

A. (a) `5/2+1/sqrt e`
B. (b) `3/2+1/sqrt e`
C. (c) `1/2+1/sqrt e`
D. (d) `3/2+sqrt e`

Answer

(b)

Explanation

Given differential equation is
`y^(2) dx+(x-1/y)dy =0`
`rArr dx/dy+1/y^(2)x=1/y^(3),` which is the linear differential
equation of the form `dx/dy+ Px = Q.`
Here, `P=1/y^(2) and Q=1/y^(3)`
Now, `IF=e^(int1/y^(2)dy)=e^(-1/y)`
`therefore` The solution of linear differential equation is
`x.(IF)intQ(IF)dy+c`
`rArr x e^(-1//y)=int 1/y^(3) e^(-1//y)dy+C`
`therefore x e^(-1//y) =int(-t) e^(t) dt+C [therefore let -1/y=trArr+1/y^(2)dy=dt]`
`=-te^(t)+int e^(t)dt+C` [integration by parts]
`=-te^(t)+ e^(t)+C`
`rArr x c^(-1//y)=1/ye^(-1//y)+e^(-1//y)+C`
Now, at y = 1, the value of x = 1, so
`1cdote^(1)=e^(-1) +e^(-1)+C rArrC=-1/e`
On putting the value of C, in Eq. (i), we get
`x=1/y+1-e^(1//y)/e`
So, at Y= 2, the value of `x=1/2+1-e^(1//2)/e= 3/2 - 1/sqrte`