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Question
If y = 3 cos (log x) + 4 sin (log x), prove that x2y2 + xy1 + y = 0 ?
Solution
Here,
\[y = 3 \cos\left( \log x \right) + 4 \sin\left( \log x \right)\]
\[\text { Differentiating w . r . t . x, we get }\]
\[ y_1 = - 3\sin\left( \log x \right) \times \frac{1}{x} + 4 \cos\left( \log x \right) \times \frac{1}{x}\]
\[ = \frac{- 3\sin\left( \log x \right) + 4\cos\left( \log x \right)}{x}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[ y_2 = \frac{\left( \frac{- 3\cos\left( \log x \right)}{x} - \frac{4\sin\left( \log x \right)}{x} \right) \times x - \left\{ - 3\sin\left( \log x \right) + 4\cos\left( \log x \right) \right\}}{x^2}\]
\[ \Rightarrow y_2 = \frac{- 3\cos\left( \log x \right) - 4\sin\left( \log x \right) - \left\{ - 3\sin\left( \log x \right) + 4\cos\left( \log x \right) \right\}}{x^2}\]
\[ \Rightarrow y_2 = \frac{- 3\cos\left( \log x \right) - 4\sin\left( \log x \right) - \left\{ - 3\sin\left( \log x \right) + 4\cos\left( \log x \right) \right\}}{x^2}\]
\[ \Rightarrow y_2 = \frac{- 3\cos\left( \log x \right) - 4\sin\left( \log x \right)}{x^2} - \frac{\left\{ - 3\sin\left( \log x \right) + 4\cos\left( \log x \right) \right\}}{x^2}\]
\[ \Rightarrow y_2 = \frac{- \left\{ 3\cos\left( \log x \right) + 4\sin\left( \log x \right) \right\}}{x^2} - \frac{\left\{ - 3\sin\left( \log x \right) + 4\cos\left( \log x \right) \right\}}{x^2}\]
\[ \Rightarrow y_2 = \frac{- y}{x^2} - \frac{y_1}{x}\]
\[ \Rightarrow x^2 y_2 = - y - x y_1 \]
\[ \Rightarrow x^2 y_2 + y + x y_1 = 0\]
Hence proved.
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