Question

The differential equation $2xydy=x^2+y^2+1dx$ determines

• A. A family of circles with centre on x-axis
• B. A family of circles with centre on y-axis
• C. A family of rectangular hyperbiola with centre on x-axis
• D. A family of rectangulat hyperbola with centre on y-axis

Answer 1

The correct option is C A family of rectangular hyperbiola with centre on x-axis

$2xy\frac{dy}{dx}=x^2+y^2+1\Rightarrow 2\frac{dy}{dx}y-\frac{y^2}{x}=\frac{1}{x}+x$
Put $v=y^2\Rightarrow \frac{dv}{dx}=2y\frac{dy}{dx}$
$\frac{dv}{dx}-\frac{v}{x}=\frac{1}{x}+x$
Let $\mu =e^{\int -\frac{1}{x}dx}=\frac{1}{x}$
$\frac{\frac{dv}{dx}}{x}-\frac{v}{x^2}=-\frac{-\frac{1}{x}-x}{x}$
Put $-\frac{1}{x^2}=\frac{d}{dx}\left(\frac{1}{x}\right)$
$\frac{\frac{dv}{dx}}{x}+\frac{d}{dx}\left(\frac{1}{x}\right)v=-\frac{-\frac{1}{x}-x}{x}$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dx}\left(\frac{v}{x}\right)=-\frac{-\frac{1}{x}-x}{x}\Rightarrow \int \frac{d}{dx}\left(\frac{v}{x}\right)dx=\int -\frac{-\frac{1}{x}-x}{x}dx\Rightarrow \frac{v}{x}=-\frac{1}{x}+x+c\Rightarrow v=x^2+cx-1\Rightarrow y^2=x^2+cx-1$