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

`x^(2)+xy + 4x+2y+4(dy)/(dx) -y^(2)=0, x gt 0`
`rArr (x+2)^(2)+y(x+2)=y^(2).(dx)/(dy)`
`rAr 1/(x+2)^(2) (dx)/(dy) =1/y^(2)+1/(y(x+2))`
`therefore 1/(x+2)^(2) (dx)/(dy) -1/((x+2)y)=1/y^(2)`…………..(i)
Put `1/(x+2)=t`,
`therefore -1/(x+2)^(2)(dx)/(dy)=(dt)/(dy)`
`therefore` Equation(i) reduces to
`-(dt)/(dy) -t/y=1/y^(2)`
`therefore` Equation (i) reduces to
`-(dt)/(dy) + t/y =-1/y^(2)`
Above in linear differential equation.
I.F. `1e^(int1/ydy)=y`
`therefore` Solution is
`t.y = inty(-1/y^(2))dy+C`
or `t.y=C-logy`
`therefore 1/(x+2).y=C-logy`
It passes through (1,3)
`rArr 1=C-log3`
`rArr C=1+log(3)`
`therefore` solution curve is `(y/(y+2))=1+log3-logy`
For `y=x+2`, we have `1=+log3-logy`
`therefore y=3`
So, option (1) is correct,
`(x+2)^(2)/(x+2)=1-log(y/3)`
`therefore x+1=log_(e)(3/y)`
`therefore y=3e^(-x-1)`
or `(x+2)^(2)=3e^(-x-1)`
Now curves `y=(x+2)^(2)` and `y=3e^(-x-1)` intersect as shown in the figure.

Form the figure no solution for `x gt 0`
For option (4)
`((x+3)^(2)/(x+2)-1) =-log{(x+3)^(2)/(3)}`
`rArr ((x+2)^(2)+1+2(x+2))/(x+2) -1+2log(x+3)-log3=0`
Let `g(x) =(x+2)+1/(x+2)+1+2log(x+3)-log3`
`rArr g^(')(x)=1-1/(x+2)^(2)+2/(x+3)`
`1+(2(x+2)^(2)-(x+3))/((x+3)(x+2)^(2))`
`therefore` g(x) is an increasing function.
Also, `g(0)=2+1/2+1+2log3-log3 lt 0`
So `g(x) ne0` for `x gt0`
Hence equation has no solution.