Solve the following differential equations: 1+x^2 dy+ 1+y^2 dx=0 - Vidyadip

Question

Solve the following differential equations:
$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$

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Answer

$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
$\sqrt{1+\text{x}^2}\ \text{dy}=-\sqrt{1+\text{y}^2}\ \text{dx}$
$\int\frac{\text{dy}}{\sqrt{1+\text{y}^2}}=-\int\frac{\text{dx}}{\sqrt{1+\text{x}^2}}$
$\log|\text{y}+\sqrt{1+\text{y}^2}|=-\log|\text{x}+\sqrt{1+\text{x}^2}|=\log|\text{c}|$
$(\text{y}+\sqrt{1+\text{y}^2})(\text{x}+\sqrt{1+\text{x}^2})=\text{c}$

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