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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
Question
Solve the following differential equation:$2 xydx + (x^{2} +2y^2) dy = 0$
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Answer

$\frac{dy}{dx} = \frac{- 2xy}{x^{2}+ 2y^{2}} ; \text{Let} y = vx\Rightarrow \frac{dy}{dx} = v + x \frac{dy}{dx}$$\therefore v + x \frac{dv}{dx} = \frac{- 2vx^2}{x + 2v^2x^2} = \frac{- 2v}{1 + 2v^2}$
$\therefore x \frac{dv}{dx} = \frac{- 2v - v - 2v^{3}}{1 + 2v^{2}} = -\frac{2v^{3} + 3v}{1 + 2v^{2}}$
$\therefore \frac{dx}{x} = \frac{1 + 2v^{2}}{(2v^{2} + 3)v}$
$= -\bigg[ \frac{2v^{2} + 3 - 2}{v (2v^{2} +3}\bigg] dv = \frac{-dv}{v} + \frac{2dv}{v (2v^{2} + 3)}$
$\therefore \frac{dx}{v} = \frac{-dv}{v} + \bigg[ \frac{A}{v} + \frac{Bv + C}{2v^{2} +3}\bigg]$
$A = \frac{2}{3}, B = -\frac{4}{3}, C = 0$
$\therefore \frac{dx}{x} = -\frac{dv}{v} + \frac{2}{3} \frac{dv}{v} -\frac{1}{3} \frac{4vdv}{2v^{2} + 3}$
Intergrating, we get
$\log x = -\log v + \frac{2}{3} \log v -\frac{1}{3} \log(2v^{2} + 3) + \log \text{c}$
$= -\frac{1}{3} \log v -\frac{1}{3} \log (2v^{2} + 3) + \log c$
$\log x^-3 = \log cv(2v^{2} + 3)$ or
$\frac{1}{x^{3}} = c \frac{y}{x} \bigg(\frac{2y^{2}}{x^{2}} + 3 \bigg)$ or
$= cy \bigg(\frac{2y^{2}}{x^{3}} + \frac{3x^{2}}{x^{3}}\bigg)$ or
$cy (2y^{2} + 3x^{2} = 1$

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