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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$
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Answer

Here, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\sqrt{\frac{\text{y}^2}{\text{x}^2}-1}$ 
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{\text{x}}-\sqrt{\frac{\text{v}^2\text{x}^2}{\text{x}^2}-1}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}-\sqrt{\text{v}^2-1}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\sqrt{\text{v}^2-1}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\sqrt{\text{v}^2-1}$
$\int\frac{\text{dv}}{\sqrt{\text{v}^2-1}}=-\int\frac{\text{dx}}{\text{x}}$
$\log\big|\text{v}+\sqrt{\text{v}^2-1}\big|=-\log|\text{x}|+\log\text{C}$
$\Big(\frac{\text{y}}{\text{x}}+\sqrt{\text{v}^2-1}\Big)=\frac{\text{C}}{\text{x}}$
$\text{y}+\sqrt{\text{y}^2-\text{x}^2}=\text{C}$

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