QJEEAdv16Maf5d8: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Maf5d8 Solution: A solution curve of the differential equation passes through…

Options (A) and (B): The given differential equation is

\left[x^2+4 x+4+y(x+2)\right] \frac{d y}{d x}-y^2=0 \quad(x>0)

which is further simplified as follows:

$\left[(x+2)^2+y(x+2)\right] \frac{d y}{d x}-y^2=0$

Substituting $x+2=t$ , we get

$\frac{d x}{d y}=\frac{d t}{d y}$

Now,

$(x+2)^2+y(x+2)-y^2 \frac{d x}{d y}=0$

That is,

$\begin{aligned}&t^2+y t-y^2 \frac{d t}{d y} =0 \\\\&y^2 \frac{d t}{d y}-y t-t^2 =0 \\\\&\frac{1}{t^2} \frac{d t}{d y}-\frac{1}{y t} =\frac{1}{y^2}\end{aligned}$

Let $\frac{1}{t}=z$ ; therefore,

$\frac{d t}{d y}\left(-\frac{1}{t^2}\right)=\frac{d z}{d y}$

Now,

$\begin{aligned} \frac{-d z}{d y}-\frac{z}{y} & =\frac{1}{y^2} \Rightarrow \frac{d z}{d y}+\frac{z}{y}=\frac{-1}{y^2} \\\\ \Rightarrow d(z y) & =\int-\frac{1}{y} d y \\\\ \Rightarrow z y & =-\ln |y|+\ln c \\\\ \Rightarrow \frac{y}{t} & =-\ln |y|+\ln c\end{aligned}$

$\Rightarrow \frac{y}{(x+2)}=-\ln |y|+\ln c ~~~~~...(1)$

which passes through the point $(1,3)$ . Therefore, from Eq. (1), we get

$\begin{aligned}&\frac{z}{3} =-\ln 3+c \Rightarrow c=\ln 3 e \\\\&\frac{y}{x+2} =-\ln |y|+\ln 3 e=\ln \left(\frac{3 e}{|y|}\right) \\\\&\frac{3 e}{|y|} =e^{y /(x+2)} \\\\&3 e =|y| e^{y /(x+2)}\end{aligned}$

Substituting $y=(x+2)$ , we get $3 e=|x+2| e^1$

$|x+2|=3 \Rightarrow x+2=-3,3 \Rightarrow x=-5,1$

Therefore, $x=1$ (since $x \neq-5)$ .


That is, the solution curve intersects $y=(x+2)$ exactly at one point and not at two points. Therefore, option (A) is correct and option (B) is incorrect.

Option (C): We have

$\frac{3 e}{\left|(x+2)^2\right|}=e^{(x+2)}$

which meets at two points for x<0 and for x>0 , there is no intersection point.

Hence, option $(\mathrm{C})$ is incorrect.

Option (D): We have

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

Therefore, there is no intersection point for x>0 . Hence option (D) is correct.