Question

The solution of differential equation $4 x y \frac{d y}{d x}=\frac{3(1+x)^2\left(1+y^2\right)}{\left(1+x^2\right)}$ is
(where $C$ is constant of integration)

• A. $\ln (1+ y )=\ln x +2 \tan x + C$
• B. $\ln \left(1+ y ^2\right)=3 \ln \frac{1}{ x }+6 \tan ^{-1} x + C$
• C. $2 \ln \left(1+y^2\right)=3 \ln x+6 \tan ^{-1} x+C$
• D. $2 \ln \left(1+ y ^2\right)=\ln x +2 \tan ^{-1} x + C$

Answer 1

Answer: (C)

Given equation can be written as $\int \frac{4 y}{1+y^2} d y=\int \frac{3(1+x)^2}{x\left(1+x^2\right)} d x$
$\Rightarrow 2 \ln \left(1+ y ^2\right) =3 \int\left(\frac{1}{ x }+\frac{2}{1+ x ^2}\right) dx $
$=3\left(\ln x +2 \tan ^{-1} x \right)+ C$