Question

Find the solution of $xy\frac{dy}{dx}=\frac{1+y^2}{1+x^2}(1+x+x^2)$.

• A. $\frac{1}{2}\log (1+y^2)=\log x+{\tan }^{-1}x+c$
• B. $\frac{1}{4}\log (1+y^2)=\log x+{\tan }^{-1}x+c$
• C. $\frac{1}{2}\log (1-y^2)=\log x+{\tan }^{-1}x+c$
• D. $\frac{1}{4}\log (1-y^2)=\log x+{\tan }^{-1}x+c$

Answer 1

The correct option is A $\frac{1}{2}\log (1+y^2)=\log x+{\tan }^{-1}x+c$

Given, $xy\frac{dy}{dx}=\frac{1+y^2}{1+x^2}(1+x+x^2)$

or $\int \frac{ydy}{1+y^2}=\int \frac{1}{x}.\frac{1+x^2+x}{1+x^2}=\int \frac{1}{x}(1+\frac{x}{1+x^2})dx=\int \frac{1}{x}+\frac{1}{1+x^2}dx$

$\int \frac{ydy}{1+y^2}=\int \frac{1}{x}+\frac{1}{1+x^2}dx$

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