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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find the particular solution of the differential equation $\text{(x - y)} \frac{\text{dy}}{\text{dx}} = \text{(x + 2y),}$ given that y = 0 when x = 1.
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Answer

$\frac{\text{dy}}{\text{dx}} = \frac{\text{x + 2y}}{\text{x - y}} = \frac{1 + \frac{\text{2y}}{\text{x}}}{1 - \frac{\text{y}}{\text{x}}}$
$\frac{\text{y}}{\text{x}} = \text{v} \Rightarrow \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}} \text{ }\text{ } \therefore \text{v + x} \frac{\text{dv}}{\text{dx}} = \frac{\text{1 + 2v}}{\text{1 -v}}$
$\Rightarrow \text{x} \frac{\text{dv}}{\text{dx}} = -\frac{\text{1 + 2v - v + v}^{2}}{\text{v - 1}} \Rightarrow \int \frac{\text{v - 1}}{\text{v}^{2} + \text{v + 1}} \text{dv} = - \frac{\text{dx}}{\text{x}}$
$\Rightarrow \int\frac{\text{2v + 1 - 3}}{\text{v}^{2} + \text{v + 1}} \text{dv} = \int - \frac{2}{\text{x}} \text{dx} \Rightarrow \int \frac{\text{2v + 1}}{\text{v}^{2} + \text{v + 1}} \text{dv - 3} \int \frac{1}{{\bigg(\text{v} + \frac{1}{2}\bigg)^{2} + \bigg(\frac{\sqrt{3}}{2}\bigg)^{2}}} = -\int \frac{2}{\text{x}} \text{dx}$
$\Rightarrow \log|\text{v}^{2} + \text{v} + 1| - 3. \frac{2}{\sqrt{3}} \tan^{-1} \bigg(\frac{\text{2v + 1}}{\sqrt{3}}\bigg) = \log |\text{x}|^{2} + \text{c}$
$\Rightarrow \log|\text{y}^{2} + \text{xy + x}^{2}| -2\sqrt{3}\tan^{-1} \bigg(\frac{\text{2y + x}}{\sqrt{3}\text{x}}\bigg) = \text{c}$
$\text{x = 1, y = 0} \Rightarrow \text{c} = -2\sqrt{3}. \frac{\pi}{6} = -\frac{\sqrt{3}}{3} \pi$
$\therefore \text{ } \log|\text{y}^{2} + \text{xy + x}^{2}| - 2\sqrt{3} \tan^{-1} \bigg(\frac{\text{2y + x}}{\sqrt{3x}}\bigg) + \frac{\sqrt{3}}{3} \pi = 0$

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