The particular solution of the differential equation $\left(1+y^2\right) d x-x y d y=0, y(1)=0$ represents
The particular solution of the differential equation $\left(1+y^2\right) d x-x y d y=0, y(1)=0$ represents
- a circle
- a part of parabola
- a part of ellipse
- a part of hyperbola
Solution
$\left(1+y^2\right) \mathrm{dx}-x y d y=0$
$\Rightarrow \frac{d x}{x}-\frac{y}{1+y^2} d y=0$
Integrating both sides:
$\begin{aligned}
& \ln x-\frac{1}{2} \ln \left(1+y^2\right)=\ln c \\
& \Rightarrow \frac{x}{\sqrt{1+y^2}}=c \Rightarrow x=\sqrt[c]{1+y^2}...(i) \\
& \because y(1)=0 \\
& \therefore 1=\sqrt[c]{1+0} \Rightarrow c=1
\end{aligned}$
From equation (i), $x=\sqrt{1+y^2} \Rightarrow x^2=1+y^2$ $\Rightarrow x^2-y^2=1$ which represents a hyperbola.
Asked in: AP EAMCET 2023 (17 May Shift 1)
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