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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
If $x = \tan \Bigg(\frac{1}{\text{a}}\log \text{y}\Bigg),$show that
$(1+\text{x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+(\text{2x - a})\frac{\text{dy}}{\text{dx}}=0$.
✓

Answer

$x = \tan \Bigg(\frac{1}{\text{a}}\log \text{y}\Bigg),$ $\Rightarrow$ $\log y = a \tan^{-1}x \therefore$ $\text{y = e}^{\text{a}\cdot\text{tan}^{-1}\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{a}\cdot\text{tan}^{-1}\text{x}}\cdot\frac{\text{a}}{\text{1 + x}^{2}}$
$\Rightarrow(1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}=\text{a}\cdot\text{y}$
differentiating again w.r.t., x we get
$ \text{(1 + x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{2x}.\frac{\text{dy}}{\text{dx}}=\text{a}\cdot\frac{\text{dy}}{\text{dx}}$
$ \text{(1 + x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{(2x - a)}\frac{\text{dy}}{\text{dx}}=0$.

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