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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
$\text{If y}=\text{e}^{\text{a}\cos^{-1}\text{x}},-1\leq\text{x}\leq1,\ \text{show that}(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0.$
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Answer

it is given that, $\text{y}=\text{e}^{\text{a}\cos^{-1}\text{x}}$
Taking logarithm on both the sides we obtain
$\log\text{y}=\text{a}\cos^{-1}\text{x}\log\text{e}$
$\log\text{y}=\text{a}\cos^{-1}\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}\times\frac{-1}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{ay}}{\sqrt{1-\text{x}^2}}$$$
By squaring both the sides, we obtain
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{a}^2\text{y}^2}{1-\text{x}^2}$
$\Rightarrow\ (1-\text{x}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{a}^2\text{y}^2$
$(1-\text{x}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{a}^2\text{y}^2$
Again differentiating both sides with respect to x, we obtain
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\frac{\text{d}}{\text{dx}}(1-\text{x}^2)+(1-\text{x}^2)\times\frac{\text{d}}{\text{dx}}\Big[\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\text{a}^2\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow\ \Big(\frac{\text{dy}}{\text{dx}}\Big)^2(-2\text{x})+(1-\text{x}^2)\times2\frac{\text{dy}}{\text{dx}}.\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a}^2.2\text{y}.\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ -\text{x}\frac{\text{dy}}{\text{dx}}+(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a}^2.\text{y}\ \Big[\frac{\text{dy}}{\text{dx}}\neq0\Big]$
$\Rightarrow\ (1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0$
Hence, proved.

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