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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
$\text{If (ax + b)} \text{e}^{\text{y/x}} = \text{x},\text{then show that}$
$\text{x}^{3} \bigg(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\bigg) = \bigg(\text{x}\frac{\text{dy}}{\text{dx}}- \text{y}\bigg)^{2} $
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Answer

$\frac{\text{y}}{\text{x}} = \log\text{x} - \log (\text{ax + b)}$
differentiating w.r.t. x,
$=\frac {\text{x} {\frac{\text{dy}}{\text{dx}}- \text{y}}}{\text{x}^{2}} = \frac{1}{\text{x}}-\frac{\text{a}}{\text{ax + b}}=\frac{\text{b}}{\text{x ( ax + b)}}$
$= \text{x}. \frac{\text{dy}}{\text{dx}} - \text{y} = \frac{\text{bx}}{(\text{ax + b)}}\dots\dots\dots\dots\text{(1)} $
differentiating w.r.t. x, again
$\text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \frac{\text{dy}}{\text{dx}} -\frac{\text{dy}}{\text{dx}} = \frac{(\text{ax + b) b - abx}}{(\text{ax + b)}^{2}} $
$\text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = \frac{\text{b}^{2}}{\text{(ax + b)}{2}}$
$\text{Writing}\Rightarrow \text{x}^{3}\ \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = \bigg(\frac{\text{bx}}{\text{ax + b}} \bigg)^{2}\dots\dots\dots\text{(2)}$
From (1) and (2) $\Rightarrow$
$\text{x}^{3} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = \bigg(\text{x}. \frac{\text{dy}}{\text{dx}}- \text{y}\bigg)^{2}$

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