Question

The solution of the differential equation, $x y \frac{d y}{d x}=\frac{1+y^2}{1+x^2}\left(1+x+x^2\right)$ given that when $x=1, y=0$ is

• A. $\ln \sqrt{1+y^2}=\ln x+\tan ^{-1} x-\frac{\pi}{2}$
• B. $\ln \frac{1+y^2}{x^2}=2 \tan ^{-1} x-\frac{\pi}{2}$
• C. $\ln \left(\frac{1+ y ^2}{ x ^2}\right)=\frac{\pi}{4}-2 \tan ^{-1} x$
• D. $\ln \frac{1+y^2}{x^2}=\tan ^{-1} x-\frac{\pi}{4}$

Answer 1

Answer: (B)

$\int \frac{ ydy }{1+ y ^2}=\int \frac{1+ x + x ^2}{ x \left(1+ x ^2\right)} dx ; \quad \frac{1}{2} \ln \left(1+ y ^2\right)=\tan ^{-1} x +\ln x + C$
$\ln \frac{1+y^2}{x^2}=2 \tan ^{-1} x+C$.