Question

A solution curve of the differential equation $\left(x^{2}+x y+4 x+2 y+4\right) \frac{d y}{d x}-y^{2}=0, x>0$, passes through the point $(1,3)$. Then the solution curve

• 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 1

Answer: (A), (D)

$\left((x+2)^{2}+y(x+2)\right) \frac{d y}{d x}-y^{2}=0$
$(x+2)(x+2+y) \frac{d y}{d x}=y^{2}$
$\frac{d y}{d x}=\frac{y^{2}}{(x+2)^{2}+(x+2) y}$
$\frac{d y}{d x}=\frac{1}{\left(\frac{x+2}{y}\right)^{2}+\frac{x+2}{y}}$
Let $\frac{x+2}{y}=t$
$x +2= ty$
$1=\frac{ dy }{ dx } \cdot t + y \frac{ dt }{ dx }$
$\left(1- y \frac{ dt }{ dx }\right) \frac{1}{ t }=\frac{ dy }{ dx }$
$\therefore \left(1-y \frac{d t}{d x}\right) \frac{1}{t}=\frac{1}{t(t+1)}$
$y \frac{d t}{d x}=\frac{t}{t+1}$
$\frac{x+2}{t} \frac{d t}{d x}=\frac{t}{t+1}$
$\int \frac{d x}{x+2}=\int \frac{t+1}{t^{2}} d t$
$\ell n(x+2)=\ell n t-\frac{1}{t}+C$
$\ell n \left(\frac{x+2}{t}\right)=-\frac{1}{t}+C$
$\Rightarrow \ell n y =\frac{- y }{ x +2}+ C $
$\left(\because y =\frac{ x +2}{ t }\right)$
It passes through $(1,3)$
$\ell n 3=-1+ C$
$ \Rightarrow C =1+\ln 3$
$\ell n y =\frac{- y }{ x +2}+1-\ln 3$
$\ell n \frac{ y }{3}=\frac{- y }{ x +2}+1$
This curve intersects $y = x +2$
$\ell n \frac{ x +2}{3}=0 $
$\Rightarrow x =-1$
Only one solution.
This curves intersects $y =( x +2)^{2}$
$\ell n \frac{( x +2)^{2}}{3}=- x -2+1=- x -1$
$(x+2)^{2}=3 e^{-x-1} \,\,\,$ but $x>0 \,\,\,$ No solution

$y=(x+3)^{2}$
$\ell n \frac{( x +3)^{2}}{3}=\frac{-( x +3)^{2}}{ x +2}+1 $
$=\frac{- x ^{2}-9-6 x + x +2}{ x +2}$