The curve satisfying the differential equation, $\left(x^2-y^2\right) d x+2 x y d y=0$ and passing through…
The curve satisfying the differential equation, $\left(x^2-y^2\right) d x+2 x y d y=0$ and passing through the point $(1,1)$ is
a circle of radius two
a circle of radius one
a hyperbola
an ellipse
Solution
$\left(x^2-y^2\right) d x+2 x y d y=0$
$
\Rightarrow \frac{d y}{d x}=\frac{y^2-x^2}{2 x y}
$
Let $y=v x$
$
\begin{aligned}
& \frac{d y}{d x}=v+x \frac{d v}{d x} \\
\Rightarrow & v+x \frac{d v}{d x}=\frac{v^2 x^2-x^2}{2 v x^2} \\
\Rightarrow & v+x \frac{d v}{d x}=\frac{v^2-1}{2 v} \\
\Rightarrow & x \frac{d v}{d x}=\frac{-v^2-1}{2 v} \\
\Rightarrow & \frac{2 v d v}{v^2+1}=-\frac{d x}{x}
\end{aligned}
$
After integrating, we get
$
\begin{aligned}
&\ln \left|v^2+1\right|=-\ln |x|+\ln c \\
&\frac{y^2}{x^2}+1=\frac{c}{x}
\end{aligned}
$
As curve passes through the point $(1,1)$, so $1+1=c \Rightarrow c=2$
$x^2+y^2-2 x=0$, which is a circle of radius one