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. Does NOT intersect $y=(x+3{)}^2$
• B. Intersects $y=(x+2{)}^2$
• C. Intersects $y=x+2$ exactly at two points
• D. Intersects $y=x+2$ exactly at one point

Answer 1

Answer: (A), (D)

Intersects $y=x+2$ exactly at one point

$(x^2+xy+4x+2y+4)\frac{dy}{dx}=y^2$
${(x+2)}^2+y(x+2)=y^2\frac{dx}{dy}$
$\frac{dx}{dy}=\frac{{(x+2)}^2}{y^2}+\frac{x+2}{y}$
$\frac{1}{{(x+2)}^2}\frac{dx}{dy}-\frac{1}{y(x+2)}=\frac{1}{y^2}$
$t=\frac{-1}{x+2}$
$\frac{dt}{dy}+\frac{t}{y}=\frac{1}{y^2}$
IF = $e^{\int \frac{dy}{y}}=y$
$y.t=\int \frac{y}{y^2}dy=\ln y+c$
$\Rightarrow \frac{-y}{x+2}=\ln y+c$
It passes through $(1,3)$
$\Rightarrow c=–1–\ln 3$
$\Rightarrow \ln y=1-\frac{y}{x+2}+\ln 3$
On solving with $y=x+2$
$\ln (x+2)=\ln 3\Rightarrow x=1$ (Exactly one point)
On solving with $y=(x+2{)}^2$
$\ln (x+2{)}^2=1–(x+2)+\ln 3$
$=–1+\ln 3–x$
As graph of LHS is increasing & graph of RHS is decreasing and ${lim}_{x\to 0^{+}}\text{LHS}>{lim}_{x\to 0^{+}}\text{RHS}$
So the curves do not intersect
Also, $(x+3{)}^2>(x+2{)}^2$ for $x>0$
So, $y=(x+3{)}^2$ and $\ln y=1+\ln 3-\frac{y}{x+2}$also do not intersect