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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation:$\sqrt{\text{1 + x}^{2}+\text{y}^{2}+\text{x}^{2}\text{y}^{2}}+\text{xy}\frac{\text{dy}}{\text{dx}}=0.$
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Answer

Given differential equation can be written as
$\sqrt{\text{(1 + x}^{2})}\sqrt{(\text{1 + y}^{2})}+\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{y}}{\sqrt{\text{1 + y}^{2}}}\text{ dy}=-\frac{\sqrt{\text{1 + x}^{2}}}{\text{x}}\text{dx}$
Integrating both sides, we get
$\sqrt{\text{1 + y}^{2}}=-\int\frac{\sqrt{\text{1 + x}^{2}}}{\text{x}^{2}}\cdot\text{x dx}=-\int\frac{\text{t}^{2}\text{ dt}}{\text{t}^{2}-1}\text{where }(1+\text{x}^{2})=\text{t}^{2}$
$\Rightarrow\sqrt{\text{1 + y}^{2}}=-\int\Bigg(1+\frac{1}{\text{t}^{2}-1}\Bigg)\text{dt}=-\text{t}-\frac{1}{2}\log\frac{\text{t - 1}}{\text{t + 1}}\text{c}$
$=-\sqrt{\text{1 + x}^{2}}-\frac{1}{2}\log\Bigg|\frac{\sqrt{\text{1 + x}^{2}}-1}{\sqrt{\text{1 + x}^{2}}+1}\Bigg|+\text{c}$
OR  $\sqrt{\text{1 + y}^{2}}+\sqrt{\text{1 + x}^{2}}-\frac{1}{2}\log\Bigg|\frac{\sqrt{\text{1 + x}^{2}}-1}{\sqrt{\text{1 + x}^{2}}+1}\Bigg|=\text{c}$.

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