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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
Find A and B so that $\text{y}=\text{A}\sin3\text{x}+\text{B}\cos3\text{x}$ satisfy the equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\frac{\text{dy}}{\text{dx}}+3\text{y}=10\cos3\text{x}.$
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Answer

$\text{y}=\text{Ae}^{-kt}\cos(\text{pt}+\text{c})$ Differentiating w.r.t.x,
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\text{A}\Big\{\text{e}^{-\text{kt}}(-\sin(\text{pt}+\text{c})\times\text{p})+(\cos(\text{pt}+{c}))(-\text{re}^{-\text{kt}})\Big\}$
$\Rightarrow-\text{Ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{KAe}^{-\text{kt}}\cos(\text{pt}+\text{c})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=-\text{Ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{ky}$ Differentiating w.r.t.x, $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{Ap}\Big\{\text{e}^{-\text{kt}}(\cos(\text{pt}+\text{c})\times\text{p})+(\sin(\text{pt}+\text{c}))(\text{e}^{-\text{kt}}\times-\text{R})-\text{ky}^1\Big\}$
$=-\text{p}^2\text{y}+\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{ky}^1$
Adding & substracting $ky_1$ on RHS $\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=+\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{y}-2\text{ky}^1+\text{ky}^1$
$\frac{\text{d}^2\text{y}}{\text{dt}^2}=\text{Apke}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{p}^2\text{y}-2\text{ky}^1-\text{k}\text{ape}^{-\text{kt}}\sin(\text{pt}+\text{c})-\text{k}^2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}=-(\text{p}+\text{k}^2)\text{y}-2\text{k}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}+2\text{k}\frac{\text{dy}}{\text{dt}}+\text{n}^2\text{y}=0$

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