Question

The general solution of the differential equation $(1+y^2)dx+(1+x^2)dy=0$ is

• A. $x–y=C(1–xy)$
• B. $x–y=C(1+xy)$
• C. $x+y=C(1–xy)$
• D. $x+y=C(1+xy)$

Answer 1

The correct option is C

$x+y=C(1–xy)$

Find the general solution of the differential equation

Given, $(1+y^2)dx+(1+x^2)dy=0$

Above equation we can also write this,

$$\begin{gathered} (1+y^2)dx=-(1+x^2)dy \\ \frac{dx}{(1+x^2)}=-\frac{dy}{(1+y^2)} \end{gathered}$$

$\frac{dx}{1+x^2}+\frac{dy}{1+y^2}=0$

On integrating, we get

$\int \frac{dx}{1+x^2}+\int \frac{dy}{1+y^2}=\int 0$

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

Hence, option (C) is the correct answer.