Question

A solution curve of the differential equation $(x^2+xy+4x+2y+4)\frac{dy}{dx}-y^2=0,x>0$, 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 pints
• C. Intersects $y={(x+2)}^2$
• D. Does N0T intersect $y={(x+3)}^2$

Answer 1

Answer: (A), (D)

The correct options are
A Does N0T intersect $y={(x+3)}^2$
D Intersects $y=x+2$ exactly at one point

$\left[\right(x+2\left)\right(x+2+y\left)\right]\frac{dy}{dx}-y^2=0$. Substitute $y=(x+2)t\Rightarrow \frac{dy}{dx}=(x+2)\frac{dt}{dx}+t$
$\Rightarrow$ $(x+2{)}^2=0$ or $(1+t)\left(\right(x+2)\frac{dt}{dx}+t)-t^2=0\Rightarrow (x+2)(1+t)\frac{dt}{dx}+t=0$
$\left(\frac{1+t}{t}\right)dt=-\frac{dx}{x+2}$
$\ln t+t=-\ln (x+2)+c\Rightarrow \ln \left(\frac{y}{x+2}\right)+\left(\frac{y}{x+2}\right)=-\ln (x+2)+c$
$\ln y-\ln (x+2)+\frac{y}{x+2}=-\ln (x+2)+c\Rightarrow \ln y+\frac{y}{x+2}=c$
$\ln 3+\frac{3}{3}=+c\Rightarrow c=\ln 3+1\Rightarrow \ln y+\frac{y}{x+2}=\ln 3e$
(A) $\ln y+\frac{y}{x+2}=\ln 3e\Rightarrow \ln (x+2)+1=\ln 3+1$
$\Rightarrow$ one solution
(D) $y=(x+3{)}^2\Rightarrow \ln (x+3{)}^2+\frac{(x+3{)}^2}{x+2}=\ln 3+1$
$2\ln (x+3)+\frac{(x+2{)}^2+1+2(x+2)}{x+2}=\ln 3+1$
$g\left(x\right)=2\ln (x+3)+(x+2)+2+\frac{1}{(x+2)}-\ln 3-1$
$g^{\prime }\left(x\right)=\frac{2}{(x+3)}+1+0-\frac{1}{(x+2{)}^2}=\frac{2(x+2{)}^2-(x+3)}{(x+3)(x+2{)}^2}+1=\frac{2x^2+8x+8-x-3}{(x+3)(x+2{)}^2}+1>0$

Since $x>0$ given, and $g\left(0\right)>0$, therefore $g\left(x\right)$ will never intersect X-axis when $x>0$. Hence option (D).