Question

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

Answer 1

We have,
$$\sqrt{1+x^2}dy+\sqrt{1+y^2}dx=0$$
$$\sqrt{1+x^2}dy=-\sqrt{1+y^2}dx$$
$$\frac{1}{\sqrt{1+y^2}}dy=-\frac{1}{\sqrt{1+x^2}}dx$$
Integrating both sides, we get
$$\int \frac{1}{\sqrt{1+y^2}}dy=-\int \frac{1}{\sqrt{1+x^2}}dx$$
$$\Rightarrow \log |y+\sqrt{1+y^2}|=-\log |x+\sqrt{1+x^2}|+\log C$$
$$\Rightarrow \log |y+\sqrt{1+y^2}|+\log |x+\sqrt{1+x^2}|=\log C$$
$$\Rightarrow \log \left|(y+\sqrt{1+y^2})(x+\sqrt{1+x^2})\right|=\log C$$
$$\Rightarrow (y+\sqrt{1+y^2})(x+\sqrt{1+x^2})=C$$
$$\text{ Hence, }\log (y+\sqrt{1+y^2})(x+\sqrt{1+x^2})=\text{ C is the required differential equation .}$$