The solution of $\left(1+y^2\right) d x-x y d y=0, y(1)=0$ represents a conic. Its eccentricity is
The solution of $\left(1+y^2\right) d x-x y d y=0, y(1)=0$ represents a conic. Its eccentricity is
- 2
- $1 / \mathrm{e}$
- 1
- $\sqrt{2}$
Solution
Given $\left(1+y^2\right) d x-x y d y-0$ and $y(1)=0$.
Take, $\left(1+y^2\right) d x-x y d y=0$
$
d x\left(1+y^2\right)=x y d y
$
$
\int \frac{d x}{x}=\int \frac{y d y}{d+y^2}
$
$
\begin{aligned}
& \mathrm{ydy}=\frac{1}{2} \mathrm{dt} \\
& \log \mathrm{x}=\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}} \\
& \log \mathrm{x}=\frac{1}{2} \log \mathrm{t}+\log \mathrm{c} \\
& \log \mathrm{x}=\log (\sqrt{\mathrm{t}} \mathrm{c})
\end{aligned}
$
$
\begin{aligned}
& x=\sqrt{1+y^2} c \\
& \text { Put } x=1, y=0 \\
& 1=\sqrt{1} c \\
& c=1
\end{aligned}
$
Then, $x=\sqrt{1+y^2}$
$
\begin{aligned}
& x^2-y^2=1 \\
& a=b=1 \\
& c=\sqrt{1+1}=\sqrt{2}
\end{aligned}
$
So, $\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}=\frac{\sqrt{2}}{\mathrm{l}}=\sqrt{2}$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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