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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation: $\frac{\text{dy}}{\text{dx}} = \frac{\text{x}(\text{2y - x})}{\text{x}(\text{2y+ x)}}\text{If, y = 1 when x = 1} $
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Answer

$\frac{\text{dy}}{\text{dx}} = \frac{\text{x}(\text{2y - x})}{\text{x}(\text{2y+ x)}}$$\text{y = vx} \Rightarrow \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}}$
$\therefore \text{v + x} \frac{\text{dv}}{\text{dx}} = \frac{\text{x[2v - 1]}}{\text{x[2v + 1]}}$
$\Rightarrow x \frac{\text{dv}}{\text{dx}} = \frac{\text{2v -1}}{\text{2v + 1}} \text{- v} = \frac{\text{2v - 1 - 2v}^{2} \text{-v}}{\text{2v + 1}}$
$= - \frac{\text{2v}^{2} \text{- v + 1}}{\text{2v + 1}}$
$\frac{\text{2v + 1}}{\text{2v}^{2} \text{- v + 1}} \text{dv} = - \frac{\text{dx}}{\text{x}}$
$\frac{1}{2} \frac{\text{4v - 1 + 3}}{\text{2v}^{2}\text{ - v+ 1}} \text{dv} = \frac{\text{-dx}}{\text{x}}$
$\frac{1}{2} \frac{\text{4v - 1 + 3}}{\text{2v}^{2}\text{ - v+ 1}} \text{dv} + \frac{3}{4} \frac{\text{dv}}{\text{v}^{2}- \frac{1}{2} \text{v} + \frac{1}{2}} = -\frac{\text{dx}}{\text{x}}$
$\frac{1}{2} \log | \text{2v}^{2} \text{- v + 1}| + \frac{3}{4} \times \frac{4}{\sqrt{7}} \tan^{-1} \frac{\text{v}-\frac{1}{4}}{{\frac{\sqrt{7}}{4}}} = -\log\text{x + c}$
$\frac{1}{2} \log\bigg|\frac{\text{2y}^{2} \text{- xy} + \text{x}^{2}}{\text{x}^{2}}\bigg| + \frac{3}{\sqrt{7}} \tan^{-1} \frac{\text{4y - x}}{\sqrt{7}\text{x}} = -\log \text{x + c}$
$\text{when x = 1, y = 1} \Rightarrow \text{c}= \frac{1}{2} \log 2 + \frac{3}{\sqrt{7}} \tan^{-1} \frac{3}{\sqrt{7}} $

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