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प्रश्न
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
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उत्तर
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
\[ \Rightarrow y\sqrt{1 + x^2}dy = - x\sqrt{1 + y^2}dx\]
\[ \Rightarrow \frac{y}{\sqrt{1 + y^2}}dy = \frac{- x}{\sqrt{1 + x^2}}dx\]
\[ \Rightarrow \int\frac{y}{\sqrt{1 + y^2}}dy = - \int\frac{x}{\sqrt{1 + x^2}}dx\]
\[\text{Let }1 + y^2 = t^2\text{ and }1 + x^2 = p^2 \]
\[ \Rightarrow 2ydy = 2tdt\text{ and }2xdx = 2pdp\]
\[ \Rightarrow ydy = tdt\text{ and }xdx = pdp\]
Substituting in above equation, we get
\[ \Rightarrow \int dt = - \int dp\]
\[ \Rightarrow t = - p + C\]
\[ \Rightarrow \sqrt{1 + x^2} + \sqrt{1 + y^2} = C\]
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