A solution curve of the differential equation ≤ft(x^2+x y+4 x+2 y+4)…

The given differential equation is

$\left(x^2+x y+4 x+2 y+4\right) \frac{d y}{d x}-y^2=0, x>0$

$\Rightarrow \left\{\left(x^2+4 x+4\right)+(x y+2 y)\right\} \frac{d y}{d x}=y^2$

$\Rightarrow \left\{(x+2)^2+y(x+2)\right\} \frac{d y}{d x}=y^2$

$\Rightarrow (x+2)(x+y+2) \frac{d y}{d x}=y^2$

$\Rightarrow \frac{d y}{d x}=\frac{y^2}{(x+2)(x+2+y)}$              ......(i)

Let $x+2=X$ and $y=Y$. Then, $\frac{d y}{d x}=\frac{d y}{d Y} \frac{d Y}{d X} \frac{d X}{d x}=\frac{d Y}{d X}$. Substituting these values in (i), we get

$\frac{d Y}{d X}=\frac{Y^2}{X(X+Y)}$

$\Rightarrow \left(X^2+X Y\right) d Y=Y^2 d X$

$\Rightarrow X^2 d Y=Y^2 d X-X Y d Y$

$\Rightarrow -\frac{1}{Y} d Y=\frac{X d Y-Y d X}{X^2}$

$\Rightarrow -\frac{1}{Y} d Y=d\left(\frac{Y}{X}\right)$

On integrating, we get

$-\log |Y|=\frac{Y}{X}+C \Rightarrow-\log |y|=\frac{y}{x+2}+C$          .....(ii)

The curve given in (ii) passes through the point $(1,3)$.

∴  $-\log 3=1+C \Rightarrow C=-\log 3-1$

Substituting the value of $C$ in (ii), we obtain

$-\log |y|=\frac{y}{x+2}-\log 3-1$             .......(iii)

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

$-\log (x+2)=-\log 3 \Rightarrow x+2= \pm 3 \Rightarrow x=1$               [∵ x > 0]

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

$-2 \log |x+2|=(x+2)-\log 3-1$

$\Rightarrow 2 \log (x+2)=-x-1+\log 3$

$\Rightarrow x+1+2 \log (x+2)=\log 3$

We find that $f(x)=x+1+2 \log (x+2)$ is strictly increasing for $x>0$ and its minimum value is $f(0)=1+2 \log 2>\log 3$.

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

$-\log (x+3)^2=\frac{(x+3)^2}{x+2}-\log 3-1$

$\Rightarrow \log \frac{(x+3)^2}{3}+\frac{(x+3)^2}{x+2}-1=0$

For $x>0, \log \frac{(x+3)^2}{3}+\frac{(x+3)^2}{x+2}-1>0$. So, the above equation has no solution. Hence, $y=(x+3)^2$ does not intersect the solution curve.