Question
Solve the following differential equations:
$\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
$\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
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Answer
We have,
$\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\sqrt{(1+\text{x}^2)(1+\text{y}^2)}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{xy}\ \frac{\text{dy}}{\text{dx}}=-\sqrt{(1+\text{x}^2)(1+\text{y}^2)}$
$\Rightarrow\text{xy}\frac{\text{dy}}{\text{dx}}=-\sqrt{(1+\text{x}^2)}\sqrt{1+\text{y}^2)}$
$\Rightarrow\frac{\text{y}}{\sqrt{(1+\text{y}^2)}}\ \text{dy}=-\int\frac{\sqrt{(1+\text{x}^2)}}{\text{x}}\ \text{dx}$
$\Rightarrow\frac{\text{y}}{\sqrt{(1+\text{y}^2)}}\ \text{dy}=-\int\frac{\text{x}\sqrt{(1+\text{x}^2)}}{\text{x}^2}\ \text{dx}$
Putting 1 + y2 = t and 1 + x2 = u2 ⇒ 2y dy = dt and 2x dx = 2udu$\Rightarrow\text{y dy}=\frac{\text{dt}}{2}$
and x dx = udu$\therefore$
intregals become,$\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}=-\int\frac{\text{u}\times\text{u}}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}=-\int\frac{\text{u}^2}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}=-\int1+\frac{1}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}-\int(1)\text{du}-\int\frac{1}{\text{u}^2-1}\ \text{du}$
$\Rightarrow\sqrt{\text{t}}=-\text{u}-\frac{1}{2}\log\Big|\frac{\text{u}-1}{\text{u}+1}\Big|+\text{C}$
$\Rightarrow\sqrt{1+\text{y}^2}=-\sqrt{1+\text{x}^2}-\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|+\text{C}$
$\Rightarrow\sqrt{1+\text{y}^2}+\sqrt{1+\text{x}^2}+\frac{1}{2}\log\Big|\frac{\sqrt{1+\text{x}^2}-1}{\sqrt{1+\text{x}^2}+1}\Big|=\text{C}$
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