Question

(1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0

Answer 1

We have, 
\[\left( 1 + x \right)\left( 1 + y^2 \right) dx + \left( 1 + y \right)\left( 1 + x^2 \right)dy = 0\]
\[ \Rightarrow \left( 1 + x \right)\left( 1 + y^2 \right) dx = - \left( 1 + y \right)\left( 1 + x^2 \right)dy\]
\[ \Rightarrow \frac{1 + x}{1 + x^2}dx = - \frac{1 + y}{1 + y^2}dy\]
Integarting both sides, we get
\[\int\frac{1 + x}{1 + x^2}dx = - \int\frac{1 + y}{1 + y^2}dy\]
\[ \Rightarrow \int\frac{1}{1 + x^2}dx + \int\frac{x}{1 + x^2}dx = - \int\frac{1}{1 + y^2}dy - \int\frac{y}{1 + y^2}dy\]
\[\text{ Substituting }1 + x^2 = t \text{ in the second integral of LHS and }1 + y^2 = u\text{ in the second integral of RHS, we get }\]
\[2x dx = dt\text{ and }2ydy = du\]
\[ \therefore \int\frac{1}{1 + x^2}dx + \frac{1}{2}\int\frac{1}{t}dt = - \int\frac{1}{1 + y^2}dy - \frac{1}{2}\int\frac{1}{u}du\]
\[ \Rightarrow \tan^{- 1} x + \frac{1}{2}\log \left| t \right| = - \tan^{- 1} y - \frac{1}{2}\log \left| u \right| + C\]
\[ \Rightarrow \tan^{- 1} x + \frac{1}{2}\log \left| 1 + x^2 \right| = - \tan^{- 1} y - \frac{1}{2}\log \left| 1 + y^2 \right| + C\]
\[ \Rightarrow \tan^{- 1} x + \tan^{- 1} y + \frac{1}{2}\log \left| 1 + x^2 \right| + \frac{1}{2}\log \left| 1 + y^2 \right| = C\]
\[ \Rightarrow \tan^{- 1} x + \tan^{- 1} y + \frac{1}{2}\log \left| \left( 1 + x^2 \right)\left( 1 + y^2 \right) \right| = C\]
\[\text{ Hence, }\tan^{- 1} x + \tan^{- 1} y + \frac{1}{2}\log \left| \left( 1 + x^2 \right)\left( 1 + y^2 \right) \right| =\text{ C is the required solution }.\]