Q41 / 120 · VITEEE 2012 · Mathematics
The solution of the differential equation $\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}$ is
- Correct answerA$y\left(1+x^{2}\right)=C+\tan ^{-1} x$
- B$\frac{\mathrm{y}}{1+\mathrm{x}^{2}}=\mathrm{C}+\tan ^{-1} \mathrm{x}$
- C$y \log \left(1+x^{2}\right)=C+\tan ^{-1} x$
- D$y\left(1+x^{2}\right)=C+\sin ^{-1} x$
Solution
$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2 \mathrm{yx}}{1+\mathrm{x}^{2}}=\frac{1}{\left(1+\mathrm{x}^{2}\right)^{2}}$
which is a linear differ ential equation.
Here, $\mathrm{P}=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}, \mathrm{Q}=\frac{1}{\left(1+\mathrm{x}^{2}\right)^{2}}$
Now, IF = $\mathrm{e}^{\int P d x}$ $=\int_{\mathrm{e}^{1+\mathrm{x}^{2}}}^{\int \frac{2 \mathrm{x}}{1+\mathrm{dx}}}=\mathrm{e}^{\log \left(1+\mathrm{x}^{2}\right)}=\left(1+\mathrm{x}^{2}\right)$
$\therefore$ Solution of differential equation is
$$
\begin{array}{l}
\text { y. }\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right) d x+C \\
\Rightarrow y\left(1+x^{2}\right)=\int \frac{1}{1+x^{2}} d x+C \\
\Rightarrow y\left(1+x^{2}\right)=\tan ^{-1} x+C
\end{array}
$$
Hence, the correct answer is option: 'A'.