The particular solution of the differential equation $\left(1+y^2\right) \mathrm{d} x-x y \mathrm{~d} y=0$…
The particular solution of the differential equation $\left(1+y^2\right) \mathrm{d} x-x y \mathrm{~d} y=0$ at $x=1, y=0$, represents
- circle
- pair of straight lines
- hyperbola
- ellipse
Solution
$\begin{array}{ll}
& \left(1+y^2\right) \mathrm{d} x-x y \mathrm{~d} y=0 \\
\therefore \quad & \left(1+y^2\right) \mathrm{d} x=x y \mathrm{~d} y \\
\therefore \quad & \frac{1}{x} \mathrm{~d} x=\frac{y \mathrm{~d} y}{1+y^2}
\end{array}$
Integrating both sides, we get
$\begin{aligned}
& \int \frac{1}{x} \mathrm{~d} x=\int \frac{y}{1+y^2} \mathrm{~d} y \\
& \log x=\frac{1}{2} \log \left(1+y^2\right)+\mathrm{c}
\end{aligned}$
$\text { At } x=1, y=0...[Given]$
$\begin{array}{ll}
\therefore & 0=\frac{1}{2} \log (1+0)+\mathrm{c} \\
\therefore & \quad \mathrm{c}=0 \\
\therefore & \log x=\frac{1}{2} \log \left(1+y^2\right) \\
\therefore & x^2=1+y^2 \\
\therefore & x^2-y^2=1,
\end{array}$
Which is a rectangular hyperbola.
Asked in: MHT CET 2023 (09 May Shift 1)
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