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

Options

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

Answer

(a,d)

Explanation

Given, `(x^(2)+xy+4x+2y+4)dy/dx-y^(2)=0`
` rArr [(x^(2)+4x+4)+y(x+2)]dy/dx-y^(2)=0`
`rArr [(x+2)^2+y(x+2)]dy/dx-y^(2)=0`
`Put x+2= X and y=Y, then`
`(X^(2) +XY)(dY)/(dx)-Y^(2)=0`
`rArr X^(2)d\Y+XYdY-Y^(2) dX=0`
`rArr X^(2)dY+Y(XdY-YdX)=0`
`rArr (dY)/Y+(XdY-YdX)/X^(2)`
`rArr d(logabs(Y))=d(Y/X)`
On integrating both sides, we get
`-log abs(Y)=Y/X+ C, where x+2=Xand y=Y`
`-log abs(Y)=y/(x+2)+C …(i)`
Since, it passes through the point, (1,3).
`therefore -log 3=1+C`
`rArr C=-1-log 3=1(loge+log3)`
`=-log 3e`
`therefore` Eq. (i) becomes
`log abs(y)+y/(x+2)-log (3e)=0`
`rArr log (abs(y)/(3e))+y/(x+2)=0 ...(ii) `
Now, to check option (a),y=x+2 interescts the curve.
`rArr log (abs(x+2)/(3e))+(x+2)/(x+2)=0 rArr log (abs(x+2)/(3e))=-1`
`rArr abs(x+2)/(3e)= e^(-1)=1/e`
`therefore x=1,-5(rejected),as x gt0`
`therefore x=1` only one solution.
Thus, (a) is the correct answer.
to check option (c), we have
`y=(x+2)^(2)and log (abs(y)/(3e)) + y/(x+2)=0`
`rArr log [(abs(x+2)^2)/(3e)]+(x+2)^(2)/(x+2)=0 rArr log [(abs(x+2)^2)/(3e)]=-(x+2)`
` rArr (x+2)^(2)/(3e)=e^((x+2))or (x+2)^(2)cdote^(x+2)=3e rArr e^(x+2)=-(3e)/((x+2)^(2))`
Clearly, they have no solution. ltbegt To check option (d), `y-(x|3)^(2)`
i.e. ` log [(abs(x+3)^2)/(3e)]+(x+3)^(2)/((x+2))=0 `

To check the number of solutions.
Let `g(x) =rlog (x+3)+(x=3)^(2)/((x+2))-log (3e) `
`therefore g'(x) = 2/(x+3) =(((x+2) cdot 2 (x+3)-(x+3)^(2))/(x+2)^(2))-0`
`=2/(x+3)+((x+3)(x+1))/(x+2)^(2)`
Clearly, when` x gt 0,` then, `g'(x)gt0`
`therefore` g(x) is increasing, when `x gt 0.`
Thus, when `xgt 0, then g(x)gt g(0)`
`g(x) gt log (3/e)+9/4gt0`