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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{x}=\sin\text{t},\text{y}=\sin\text{pt}$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{y}=0$
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Answer

Here,
$\text{x}=\sin\text{t},\text{y}=\sin\text{pt}$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{dt}}=\cos\text{t}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{o}\cos\text{pt}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{p}\cos\text{pt}}{\cos\text{t}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{o}^2\sin\text{pt}\cos\text{t}+\text{p}\cos\text{pt}\sin\text{t}}{\cos^3\text{t}}\times\frac{\text{dt}}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{o}^2\sin\text{pt}\cos\text{t}+\text{p}\cos\text{pt}\sin\text{t}}{\cos^3\text{t}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{p}\sin\text{pt}\cos\text{t}}{\cos^3\text{t}}+\frac{\text{p}\cos\text{pt}\sin\text{t}}{\cos^3\text{t}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{p}^2\text{y}}{\cos^2\text{t}}+\frac{\text{x}\frac{\text{dy}}{\text{dx}}}{\cos^3\text{t}}$
$\Rightarrow\cos^2\text{t}\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{p}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\sin^2\text{t})\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{p}^2\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{y}=0$

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