Question

The solution of $\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}$ satisfying $y\left(1\right)=1$ is given by

• A. A system of hyperbolas
• B. A system of circles
• C. $y^2=x(1+x)-1$
• D. $(x-2{)}^2+(y-3{)}^2=5$

Answer 1

Answer: (A), (C)

The correct options are
A A system of hyperbolas
C $y^2=x(1+x)-1$

Given: $\frac{dy}{dx}=\frac{x^2+y^2+1}{2xy}$; $y\left(1\right)=1$
$2xydy=x^{2}dx+y^{2}dx+dx$
$\Rightarrow x\left(2ydy\right)=x^{2}dx+y^{2}dx+dx$
$\Rightarrow x\left(2ydy\right)-y^{2}dx=(x^2+1)dx$
Divide by $x^2$ on both sides
$\Rightarrow \frac{x\left(2ydy\right)-y^{2}dx}{x^2}=(1+\frac{1}{x^2})dx$
$\Rightarrow d\left(\frac{y^2}{x}\right)=(1+\frac{1}{x^2})dx$
$\Rightarrow \frac{y^2}{x}=(x-\frac{1}{x})+c$
$\because y\left(1\right)=1$
$\Rightarrow 1=1-1+c\Rightarrow c=1$
$\Rightarrow \frac{y^2}{x}=x-\frac{1}{x}+1$
$\Rightarrow y^2=x^2-1+x$

$\Rightarrow y^2=x(x+1)-1$

$\Rightarrow x^2-y^2+x-1=0$
$\Rightarrow x^2+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+2\times \frac{1}{2}\times x+-y^2-1=0$
$\Rightarrow {(x+\frac{1}{2})}^2-y^2=\frac{5}{4}$