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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation: $(\text{x}^{2} - \text{y}^{2}) \text{dx} + \text{2xy dy =0}$ given that $y = 1$ when $x = 1$
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Answer

$(\text{x}^{2} - \text{y}^{2}) \text{dx} + \text{2xy dy =0}$$\frac{\text{dy}}{\text{dx}} = \frac{\text{y}^{2} - \text{x}^{2}}{\text{2xy}}$
This is a homogeneous differential equation
$\text{Let y = vx} \Rightarrow \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}}$
$= \text{v + x} \frac{\text{dv}}{\text{dx}} = \frac{\text x^{2}(\text v^{2}-1)}{2\text v \text x^{2}} = \frac{\text v^{2} - 1}{2\text v}$
$\text{x}\frac{\text{dv}}{\text{dx}} = \frac{\text{v}^{2} - 1 - 2\text{v}^{2}}{2\text v} = \frac{1 + \text{v}^{2}}{2\text{v}}$
$\Rightarrow \frac{2\text{v}}{1 + \text{v}^{2}} \text{dv} = - \frac{\text{dx}}{\text{x}}$
$\log | 1 + \text{v}^{2}| = -\log| \text{x}| + \log \text{c} = \log\frac{\text{c}}{\text{x}}$
$1 + \text v^{2} = \frac{\text{c}}{\text{x}} \Rightarrow 1 + \frac{\text{y}^{2}}{\text{x}^{2}}= \frac{\text{c}}{\text{x}}$
$\Rightarrow \text{x}^{2} + \text{y}^{2} = \text{cx}$
$\text{when x = 1, y = 1,} \Rightarrow \text{c} = 2$
$\therefore \text{x}^{2} + \text{y}^{2} = \text{2x}$

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