If `x dy =y (dx+y dy), y(1) =1`and `Y(x)gt 0`. Then, `y (-3)
is epual to

Options

A. 3 only
B. `-1` only
C. Both -1 and 3
D. Neither -1 or 3

Answer

A

Explanation

Given xd= ydx+ ` y^(2) dy`
` Rightarrow 1 = y/x (dx)/(dy) + Y^(2)/x`
` Rightarrow (dx)/(dy) +y = x/y`
` Rightarrow (dy)/(dx) - x/y =-y`
` P= -1/y ,Q =-y`
`IF = e^(int pdy) = e^(int 1/ydy) = e^(-logy) = 1/y`
Now, solution of d.E.
x(I.F) = ` int (Q.I.F) dy`
` x/y = int 1/y (-y) dy +C`
`Rightarrow x/y = int -1 dy +C`
` Rightarrow x/y = -y +C`
y(1) =1
` 1/1 = -1 +C Rightarrow C =2`
` Rightarrow x/y = - y+2 Rightarrow x = -y^(2) +2y`
` Rightarrow y(-3) Rightarrow -3 =-y^(2) +2y`
`Rightarrow y^(2) -2 y -3=0`
` Rightarrow y = ( +2 +- sqrt(4+12))/2 = (2+-4)/2`
`Rightarrow y=3 , -1`
since y gt 0 so y =3