Q105 / 180 · COMEDK 2018 · Mathematics
The particular solution of $\frac{y}{x} \frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$, when $x=1, y=2$ is
- A$5\left(1+y^{2}\right)=2\left(1+x^{2}\right)$
- Correct answerB$2\left(1+y^{2}\right)=5\left(1+x^{2}\right)$
- C$5\left(1+y^{2}\right)=\left(1+x^{2}\right)$
- D$\left(1+y^{2}\right)=2\left(1+x^{2}\right)$
Solution
We have, $\quad \frac{y}{x} \frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$
$$
\begin{aligned}
&\frac{y d y}{1+y^{2}}=\frac{x d x}{1+x^{2}} \\
&\frac{2 y d y}{1+y^{2}}=\frac{2 x d x}{1+x^{2}}
\end{aligned}
$$
Integrating both sides, we get
$$
\int \frac{2 y d y}{1+y^{2}}=\int \frac{2 x d x}{1+x^{2}}
$$
$$
\begin{array}{r}
\ln \left(1+y^{2}\right)=\ln \left(1+x^{2}\right)+\ln c \\
1+y^{2}=c\left(1+x^{2}\right)
\end{array}
$$
Now, $y=2$ at $x=1$
So, $\quad 5=c(2)$
$$
2 c=5
$$
Therefore, $2\left(1+y^{2}\right)=5\left(1+x^{2}\right)$
Hence, the correct answer is option: 'B'.