Question

For the differential equation $2xydy-(x^2+y^2+1)dx=0.$

• A. $y=\sqrt{x^2+cx-1}$
• B. $y=\sqrt{x^2+cx+1}$
• C. $y=-\sqrt{x^2+cx-1}$
• D. $y=-\sqrt{x^2+cx+1}$

Answer 1

Answer: (A), (C)

$y=-\sqrt{x^2+cx-1}$
$y=\sqrt{x^2+cx-1}$

$2xydy-(x^2+y^2+1)dx=0$

$\displaystyle \Rightarrow 2y\frac { dy }{ dx } -\frac { { y }^{ 2 } }{ x } =\frac { 1 }{ x } +x$

Substitute $v=y^2\Rightarrow dv=2ydy$

$\therefore \frac{dv}{dx}-\frac{v}{x}=\frac{1}{x}+x$ ...(1)

Here $P=-\frac{1}{x}\Rightarrow \int Pdx=-\int \frac{1}{x}dx=\log \frac{1}{x}$

$\therefore I.F.=e^{\log \frac{1}{x}}=\frac{1}{x}$

Multiplying (1) by $I.F.$ we get

$\frac{1}{x}\frac{dv}{dx}-\frac{v}{x^2}=-\frac{-\frac{1}{x}-x}{x}$

Integrating both sides w.r.t $x$ we get

$\frac{v}{x}=-\int \frac{-\frac{1}{x}-x}{x}dx+c=x+c-\frac{1}{x}$

$\Rightarrow v=x^2+cx-1$

$\Rightarrow y^2=x^2+cx-1$

$\Rightarrow y=\pm \sqrt{x^2+cx-1}$