Question

A solution 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 points
• C. intersects $y=(x+2{)}^2$
• D. does NOT intersect $y=(x+3{)}^2$

Answer 1

Answer: (A), (D)

does NOT intersect $y=(x+3{)}^2$

$(x^2+xy+4x+2y+4)\frac{dy}{dx}-y^2=0,x>0$
$(x+2)(x+y+2)\frac{dy}{dx}=y^2$

Let $x+2=t$
$\Rightarrow t(t+y)\frac{dy}{dt}=y^2$
$\Rightarrow \frac{dy}{dt}=\frac{(y/t{)}^2}{1+y/t}$

Let $\frac{y}{t}=k$
$\Rightarrow k+t\frac{dk}{dt}=\frac{k^2}{1+k}$
$\Rightarrow \left(\frac{k+1}{k}\right)dk+\frac{dt}{t}=0$
$\Rightarrow k+\ln |kt|=c$
$\frac{y}{x+2}+\ln |y|=c$

Now, Given that the curve passes through the point $(1,3)$.
$\Rightarrow c=\ln 3e$
$\frac{y}{x+2}+\ln |y|=\ln 3e$

Option $1\&2$:
For the curve $y=x+2$
$1+\ln (x+2)=\ln 3e$
LHS is an increasing function & RHS is a constant
$\text{Min}(1+\ln (x+2)=\ln 2e<\ln 3e$
So, there will be only one intersection point of the curves

Option $3$:
For the curve $y=(x+2{)}^2$
$x+2+2\ln (x+2)=\ln 3e$
LHS is an increasing function & RHS is a constant
$\text{Min}(x+2+2\ln (x+2\left)\right)=2+2\ln 2>\ln 3e$
So, the curve won't intersect eachother

Option $4$:
For the curve $y=(x+2{)}^3$
$(x+2{)}^2+3\ln (x+2)=\ln 3e$
LHS is an increasing function & RHS is a constant
$\text{Min}\left(\right(x+2{)}^2+3\ln (x+2)=4+3\ln 2>\ln 3e$
So, the curve won't intersect eachother