Question

A solution curve of the differential equation $\left(x^2+xy+4x+2y+4\right)\frac{dy}{dx}-y^2=0,x>0$, 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 points
• C. intersects $y=(x+2{)}^2$
• D. does NOT intersect $y=(x+3{)}^2$

Answer 1

Answer: (A), (D)

$\left((x+2{)}^2+y(x+2)\right)\frac{dy}{dx}-y^2=0$
$(x+2)(x+2+y)\frac{dy}{dx}=y^2$
$\frac{dy}{dx}=\frac{y^2}{(x+2{)}^2+(x+2)y}$
$\frac{dy}{dx}=\frac{1}{{\left(\frac{x+2}{y}\right)}^2+\frac{x+2}{y}}$
Let $\frac{x+2}{y}=t$
$x+2=ty$
$1=\frac{dy}{dx}\cdot t+y\frac{dt}{dx}$
$\left(1-y\frac{dt}{dx}\right)\frac{1}{t}=\frac{dy}{dx}$
$\therefore \left(1-y\frac{dt}{dx}\right)\frac{1}{t}=\frac{1}{t(t+1)}$
$y\frac{dt}{dx}=\frac{t}{t+1}$
$\frac{x+2}{t}\frac{dt}{dx}=\frac{t}{t+1}$
$\int \frac{dx}{x+2}=\int \frac{t+1}{t^2}dt$
$\ell n(x+2)=\ell nt-\frac{1}{t}+C$
$\ell n\left(\frac{x+2}{t}\right)=-\frac{1}{t}+C$
$\Rightarrow \ell ny=\frac{-y}{x+2}+C$
$\left(\because y=\frac{x+2}{t}\right)$
It passes through $(1,3)$
$\ell n3=-1+C$
$\Rightarrow C=1+\ln 3$
$\ell ny=\frac{-y}{x+2}+1-\ln 3$
$\ell n\frac{y}{3}=\frac{-y}{x+2}+1$
This curve intersects $y=x+2$
$\ell n\frac{x+2}{3}=0$
$\Rightarrow x=-1$
Only one solution.
This curves intersects $y=(x+2{)}^2$
$\ell n\frac{(x+2{)}^2}{3}=-x-2+1=-x-1$
$(x+2{)}^2=3e^{-x-1}$ but $x>0$ No solution

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