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Higher Order Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives4 Marks
Question
If $\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x}),$ prove that $\text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0$
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Answer

Here,
$\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x}),$
Differentiating w.r.t.x, we get
$\text{y}_1=-3\sin(\log\text{x})\times\frac{1}{\text{x}}+4\cos(\log\text{x})\times\frac{1}{\text{x}}$
$=\frac{-3\sin(\log\text{x})+4\cos(\log\text{x})}{\text{x}}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{\Big(\frac{-3\cos(\log\text{x})}{\text{x}}-\frac{4\sin(\log\text{x})}{\text{x}}\Big)\times \text{ x}-\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})-4\sin(\log\text{x})-\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})-4\sin(\log\text{x})}{\text{x}^2}-\frac{\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-3\cos(\log\text{x})+4\sin(\log\text{x})}{\text{x}^2}-\frac{\{-3\sin(\log\text{x})+4\cos(\log\text{x})\}}{\text{x}^2}$
$\Rightarrow\text{y}_2=\frac{-\text{y}}{\text{x}^2}-\frac{\text{y}_1}{\text{x}}$
$\Rightarrow\text{x}^2\text{y}_2=-\text{y}-\text{xy}_1$
$\Rightarrow\text{x}^2\text{y}_2+\text{y}+\text{xy}_1=0$
Hence proved

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