A solution curve of the differential equation `(x^(2)+xy+4x+2y+4)(dy)/(dx)-y^(2)=0,xgt0,` passes through the point (1,3) Then, the solution curve

Options

A. intersects `y=x+2` exactly at one point
B. intersects `y=x+2` exactly at two points
C. intersects `y=(x+2)^(2)`
D. does not intersect `y=(x+3)^(2)`

Answer

A, D

Explanation

The given differential equation is
`(x^(2)+xy+4x+2y+4)(dy)/(dx)-y^(2)=0, x gt 0`
`rArr" "{(x^(2)+4x+4)+(xy+2y)}(dy)/(dx)=y^(2)`
`rArr" "{(x+2)^(2)+y(x+2)}(dy)/(dx)=y^(2)`
`rArr" "(x+2)(x+y+2)(dy)/(dx)=y^(2)`
`rArr" "(dy)/(dx)=(y^(2))/((x=2)(x+2+y))" ...(i)"`
Let `x+2=X` and `y=Y`. Then, `(dy)/(dx)=(dy)/(dY)=(dY)/(dX)(dX)/(dx)=(dY)/(dX)`.
Substituting these values in (i), we get
`(dY)/(dX)=(Y^(2))/(X(X+Y))`
`rArr" "(X^(2)+XY)dY=Y^(2)dX`
`rArr" "X^(2)dY=Y^(2)dX-XYdY`
`-(1)/(Y)dY=(XdY-YdX)/(X^(2))`
`-(1)/(Y)dY=d((Y)/(X))`
On integrating, we get
`-log|Y|-(Y)/(X)+CrArr-log|y|=(y)/(x+2)+C" ...(ii)"`
The curve given in (ii) passes through the point `(1, 3)`.
`therefore" "-log3=1+C rArr C=-log3-1`
Substituting the value of C in (ii), we obtain
`-log|y|=(y)/(x+2)-log3-1`
In order to find the points of intersection (if any) of (iii) and `y=x+2`, we solve the two equations simultaneously.
Putting `y=x+2` in (iii), we obtain
`-log(x+2)=-log3 rArr x+2 = pm 3 rArr x=1." "[because x gt0]`
Putting x = 1 in `y=x+2`, we obtain y = 3.
So, the solution curve (iii) intersects `y=x+2` exactly at one point (1,3). So, option (a) is correct but option (b) is incorrect.
Putting `y=(x+2)^(2)` in (iii), we obtain
`-2log|x+2|=(x+2)-log3-1`
`rArr" "2log(x+2)=-x-1+log3`
`rArr" "x+1+2 log(x+2)=log3`
we find that `f(x)=x+1+2 log(x+2)` is strictly increasing for `x gt0` and its minimum value is `f(0)=1+2log 2 gt log3`.
So, the above equation has no solution. So, `y=(x=2)^(2)` does not intersect the solution curve.
To find the point of intersection of the solution curve (iii) and `y=(x+3)^(2)` , we put `y=(x+3)^(2)` in (iii) to get
`-log(x+3)^(2)=((x+3)^(2))/(x+2)-log3-1`
`rArr" "log.((x+3)^(2))/(3)+((x+3)^(2))/(x+2)-1=0`
For `x gt 0, log.((x+3)^(2))/(3)+((x+3)^(2))/(x+2)-1gt 0`. So, the above equation has no solution. Hence, `y=(x+3)^(2)` does not intersect the solution curve.