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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
$\text{If y}=3\cos(\log \text{x})+4\sin(\log\text{x}),\text{ show that }\text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0$
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Answer

Given: $\text{y}=3\cos(\log\text{x})+4\sin(\log\text{x})\ \dots\text{(i)}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{y}_1=-3\sin(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}+4\cos(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\ \text{y}_1=-3\sin(\log\text{x})\frac{1}{\text{x}}+4\cos(\log\text{x})\frac{1}{\text{x}}$ $=\frac{1}{\text{x}}[-3\sin(\log\text{x})+4\cos(\log\text{x})]$
$\Rightarrow\ \text{xy}_1=-3\sin(\log\text{x})+4\cos(\log\text{x})$
Now $\frac{\text{d}}{\text{dx}}(\text{xy}_1)=-3\cos(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}-4\sin(\log\text{x})\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\ \text{x}\frac{\text{d}}{\text{dx}}(\text{y}_1)+\text{y}_1\frac{\text{d}}{\text{dx}}\text{x}$ $=-3\cos(\log\text{x})\frac{1}{\text{x}}-4\sin(\log\text{x})\frac{1}{\text{x}}$
$\Rightarrow\ \text{xy}_2+\text{y}_1=-\frac{[3\cos(\log\text{x})+4\sin(\log\text{x})]}{\text{x}}$
$\Rightarrow\ \text{x}(\text{xy}_2+\text{y}_1)=-[3\cos(\log\text{x})+4\sin(\log\text{x})]$
$\Rightarrow\ \text{x}(\text{xy}_2+\text{y}_1)=-\text{y}\ \ [\text{From eq.(i)}]$
$\Rightarrow\ \text{x}^2\text{y}_2+\text{xy}_1+\text{y}=0 \ \text{ Hence proved}.$

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