Question

The solution of the differential equation $\frac{dy}{dx}+\frac{2xy}{1+x^2}={\left(\frac{1}{1+x^2}\right)}^2$ is

• A. $y(1-x^2)={\tan }^{-1}x+c$
• B. $y(1+x^2)={\tan }^{-1}x+c$
• C. $y{(1+x^2)}^2={\tan }^{-1}x+c$
• D. $y{(1-x^2)}^2={\tan }^{-1}x+c$

Answer 1

The correct option is B

$y(1+x^2)={\tan }^{-1}x+c$

Explanation for the correct option

Given: $\frac{dy}{dx}+\frac{2xy}{1+x^2}={\left(\frac{1}{1+x^2}\right)}^2$

Let $P=\frac{2x}{1+x^2}$and $Q={\left(\frac{1}{1+x^2}\right)}^2$

Integrating factor $I=e^{\int Pdx}$

$$\begin{gathered} I=e^{\int \frac{2x}{1+x^2}dx} \\ \Rightarrow I=e^{\log (1+x^2)}\left[\int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\log \left(f\right(x\left)\right)\right] \\ \Rightarrow I=1+x^2\left[e^{\log \left(x\right)}=x\right] \end{gathered}$$

general solution$=\frac{1}{I}\int I·Q·dx$

$$\begin{gathered} \Rightarrow y=\frac{1}{1+x^2}\int \left(1+x^2\right){\left(\frac{1}{1+x^2}\right)}^2·dx \\ \Rightarrow y=\frac{1}{1+x^2}\int \left(\frac{1}{1+x^2}\right)·dx\left[\int \left(\frac{1}{1+a^2}\right)·da={\tan }^{-1}\left(a\right)+c\right] \\ \Rightarrow y(1+x^2)={\tan }^{-1}\left(x\right)+c \end{gathered}$$

Hence, option B is correct.