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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
Find $\text{y}=\text{Ae}^{-\text{kt}}\cos\text({pt}+\text{c})$prove that $\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+\text{n}^2\text{y}=0,$Where $\text{n}^2=\text{p}^2+\text{k}^2.$
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Answer

We have,
$\text{y}=\text{Ae}^{-\text{kt}}\cos\text({pt}+\text{c})...(1)$
Differentiating y with respect to t, we get
$\frac{\text{dy}}{\text{dt}}=-\text{KAe}^{-\text{kt}}\cos(\text{pt}+\text{c})-\text{PAe}^{-\text{kt}}\sin(\text{pt}+\text{c})$
$=-\text{ky}-\text{PAe}^{-\text{kt}}\sin(\text{pt}+\text{c})\ [\text{from}(1)]$
$\Rightarrow\text{pAe}^{-\text{kt}}\sin(\text{pt}+\text{c})=-\text{ky}-\frac{\text{dy}}{\text{dt}}...(2)$
Differentiating $\frac{\text{dy}}{\text{dt}}$ with respect to t, we get
$\frac{\text{d}^2\text{y}}{\text{dt}}=-\text{k}\frac{\text{dy}}{\text{dt}}+\text{pkAe}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{Ae}^{-\text{kt}}\cos(\text{pt}+\text{c})$
$=-\text{k}\frac{\text{dy}}{\text{dt}}+\text{k}\Big(-\text{ky}-\frac{\text{dy}}{\text{dt}}\Big)-\text{p}^2\text{y}\ [\text{from}(1)\ \text{and}\ (2)]$
$=-\text{k}\frac{\text{dy}}{\text{dt}}-\text{k}^2\text{y}-\text{k}\frac{\text{dy}}{\text{dt}}-\text{p}^2\text{y}$
$=-2\text{k}\frac{\text{dy}}{\text{dt}}-(\text{k}^2+\text{p}^2)\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+(\text{k}^2+\text{p}^2)\text{y}=0$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}} \text{n}^2\text{y}=0,$ where $\text{n}^2=\text{p}^2+\text{k}^2.$
Hence,
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}} \text{n}^2\text{y}=0,$ where $\text{n}^2=\text{p}^2+\text{k}^2.$

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