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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives3 Marks
Question
If $\text{y}=\text{e}^{\text{a}\cos^{-1}}\text{x}$ prove 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

Here,
$\text{y}=\text{e}^{\text{a}\cos^{-1}}\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=-\text{e}^{\text{a}\cos^{-1}}\text{x}\ \times\frac{\text{a}}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^{\text{a}\cos^{-1}}\text{x}\ \times\frac{\text{a}^2}{1-\text{x}^2}+\frac{\text{xa }\text{e}^{\text{a}\cos^{-1}}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}\times\frac{\text{a}^2}{1-\text{x}^2}-\frac{\text{x}\frac{\text{dy}}{\text{dx}}}{(1-\text{x}^2)}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{a}^2\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0$

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