A solution curve of differential equation x 2 + x y + 4 x + 2 y + 4 d y d x - y 2 = 0 ,   x >…

A solution curve of differential equation x2+xy+4x+2y+4dydx-y2=0, x>0  passes through the point (1, 3). Then the solution curve -
  1. Intersects y=x+2 exactly at one point
  2. Intersects y=x+2 exactly at two points
  3. Intersects y=x+22
  4. Does NOT intersect y=x+32

Solution

x2+xy+4x+2y+4dydx-y2=0
⇒x+22+yx+2dydx=y2
Let x+2=X, y=Y
⇒XX+YdYdX=Y2
⇒-X2dY=XYdY-Y2dX
⇒-X2dY=YXdY-YdX
⇒-dYY=XdY-YdXX2
Integrate both sides
⇒ -lnY=YX+C
Putting back X, Y in x, y terms
⇒-lny=yx+2+C
∵ it is passing through (1, 3) , - ln3=1+C
C= -1- ln3
∴ curve equation is yx+2+ lny-1- ln3=0, x>0 .....(i)
Put y=x+2 in equation (i)
then x+2x+2+ lnx+2-1- ln 3=0
x=1, -5 (reject as given x>0 )
∴ curve intersect y=x+2 at point (1, 3),(Only one point); put y=x+22 , we will get x+2+2lnx+2=1+ ln3
Clearly left hand side is an increasing function. Hence, it is always greater than 2+2ln2 (By Putting X = 0)
Therefore no solution y=x+22 put y=x+32 in equation (i)
x+32x+2+lnx+32-1-ln3=0
x+32x+2+lnx+323-1=0
∵ x>0   ⇒  x+3>x+2 and x+3>3
So x+32x+2+lnx+323>1
∴ x+32x+2+lnx+323-1=0 has no solution
⇒ curve y =x+32 does not intersect

Asked in: JEE Advanced 2016 (Paper 1)

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