The solution of (xsqrt{1 + y^2})dx + (ysqrt{1 | vedclass

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(A) Given differential equation is: $(x\sqrt{1 + y^2})dx + (y\sqrt{1 + x^2})dy = 0$Rearranging the terms to separate the variables:$(x\sqrt{1 + y^2})d

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$\sqrt{1 + x^2} + \sqrt{1 + y^2} = c$

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(A) Given differential equation is: $(x\sqrt{1 + y^2})dx + (y\sqrt{1 + x^2})dy = 0$Rearranging the terms to separate the variables:$(x\sqrt{1 + y^2})dx = -(y\sqrt{1 + x^2})dy$Dividing both sides by $\sqrt{1 + x^2}\sqrt{1 + y^2}$:$\frac{x}{\sqrt{1 + x^2}}dx = -\frac{y}{\sqrt{1 + y^2}}dy$Integrating both sides:$\int \frac{x}{\sqrt{1 + x^2}}dx = -\int \frac{y}{\sqrt{1 + y^2}}dy$Let $u = 1 + x^2$,then $du = 2xdx$,so $xdx = \frac{1}{2}du$. Similarly for $y$:$\frac{1}{2} \int u^{-1/2} du = -\frac{1}{2} \int v^{-1/2} dv$$\sqrt{u} = -\sqrt{v} + C$Substituting back the values of $u$ and $v$:$\sqrt{1 + x^2} = -\sqrt{1 + y^2} + C$$\sqrt{1 + x^2} + \sqrt{1 + y^2} = C$Thus,the correct option is $A$.

The solution of $(x\sqrt{1 + y^2})dx + (y\sqrt{1 + x^2})dy = 0$ is

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