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Question
Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].
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Solution
\[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{\sqrt{1 + x^2 + y^2 + x^2 y^2}}{xy}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{\sqrt{\left( 1 + x^2 \right) + y^2 \left( 1 + x^2 \right)}}{xy}\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{\sqrt{\left( 1 + x^2 \right)\left( 1 + y^2 \right)}}{xy}\]
\[ \Rightarrow \frac{y}{\sqrt{1 + y^2}}dy = - \frac{\sqrt{1 + x^2}}{x}dx\]
\[ \Rightarrow \left( 1 + y^2 \right)^{- \frac{1}{2}} ydy = - \frac{\sqrt{1 + x^2}}{x}dx\]
Integrating both sides, we get
\[\frac{1}{2}\int \left( 1 + y^2 \right)^{- \frac{1}{2}} 2ydy = - \int\frac{\sqrt{1 + x^2}}{x}dx\]
\[ \Rightarrow \sqrt{1 + y^2} = - \int\frac{\sqrt{1 + x^2}}{x}dx . . . . . \left( 1 \right) \left[ \int \left[ f\left( x \right) \right]^n f'\left( x \right)dx = \frac{\left[ f\left( x \right) \right]^{n + 1}}{n + 1} + C \right]\]
\[\text { Let} I_1 = \int\frac{\sqrt{1 + x^2}}{x}dx\]
Put x = tanθ
\[\Rightarrow\] dx = sec2θdθ
\[\therefore I_1 = \int\frac{\sqrt{1 + \tan^2 \theta}}{\tan\theta} \sec^2 \theta d\theta\]
\[ = \int\frac{\sqrt{\sec^2 \theta}}{\tan\theta} \sec^2 \theta d\theta\]
\[ = \int\frac{\sec^3 \theta}{\tan\theta}d\theta\]
\[ = \int\frac{1}{\sin\theta \cos^2 \theta}d\theta\]
\[ = \int\frac{\sin^2 \theta + \cos^2 \theta}{\sin\theta \cos^2 \theta}d\theta\]
\[ = \int\tan\theta\sec\theta d\theta + \int\cos ec\theta d\theta\]
\[ = \sec\theta + \log\left( \cos ec\theta - \cot\theta \right)\]
\[ = \sqrt{1 + \tan^2 \theta} + \log\left( \sqrt{1 + \frac{1}{\tan^2 \theta}} - \frac{1}{\tan\theta} \right)\]
\[\therefore I_1 = \sqrt{1 + x^2} + \log\left( \sqrt{1 + \frac{1}{x^2}} - \frac{1}{x} \right) + C . . . . . \left( 2 \right)\]
From (1) and (2), we have
\[\sqrt{1 + y^2} = \sqrt{1 + x^2} + \log\left( \sqrt{1 + \frac{1}{x^2}} - \frac{1}{x} \right) + C\]
\[ \Rightarrow \sqrt{1 + y^2} = \sqrt{1 + x^2} + \log\left( \frac{\sqrt{1 + x^2} - 1}{x} \right) + C\]
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