Find the solution of the differential equation x 1+y^ 2 dx+y 1+x^… - Vidyadip

Question

Find the solution of the differential equation $\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0.$

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Answer

$\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0$
$\Rightarrow \text{y}\sqrt{1+\text{x}^{2}}\text{dy}=-\text{x}\sqrt{1+\text{y}^{2}}\text{dx}=0$
$\Rightarrow \frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=\frac{-\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
$\Rightarrow \int\frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=-\int\frac{\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
Let $1+\text{y}^{2}=\text{t}^{2}=\text{t}^{2}, 1+\text{x}^{2}=\text{p}^{2}$
$\Rightarrow 2\text{y}\ \text{dy}=2\text{t}\ \text{dt}, 2\text{x}\ \text{dx}=2\text{p}\ \text{dp}$
$\Rightarrow \text{y}\ \text{dy}=\text{t}\ \text{dt}, \text{x}\ \text{dx}=\text{p}\ \text{dp}$
Substituting in above equation, we get
$\Rightarrow \int\text{dt}=-\int\text{dp}$
$\Rightarrow \text{t}=-\text{p}+\text{C}$
$\Rightarrow\sqrt{1+\text{x}^{2}}+\sqrt{1+\text{y}^{2}}=\text{C}$

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