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Question

If $$y = a\cos (\log x)- b\sin (\log x)$$, then the value of $$x^{2} \dfrac {d^{2}y}{dx^{2}} + x\dfrac {dy}{dx} + y$$ is


A
0
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B
1
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C
2
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D
3
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Solution

The correct option is A $$0$$
Given, $$y = \cos (\log x) - b\sin (\log x)$$
On differentiating w.r.t. $$x$$, we get
$$\dfrac {dy}{dx} = a\dfrac {[-\sin (\log x)]}{x} - \dfrac {b\cos (\log x)}{x}$$
$$= -\dfrac {[a\sin (\log x)] + b\cos (\log x)}{x}$$
$$\Rightarrow x\dfrac {dy}{dx} = -[a\sin (\log x) + b\cos (\log x)]$$
Again, on differentiating w.r.t. $$x$$, we get
$$x\dfrac {d^{2}y}{dx^{2}} + \dfrac {dy}{dx} = -\left [\dfrac {a\cos (\log x)}{x} - \dfrac {b\sin (\log x)}{x}\right ] = -\dfrac {y}{x}$$
$$\Rightarrow x^{2} \dfrac {d^{2}y}{dx^{2}} + x\dfrac {dy}{dx} + y = 0$$

Mathematics

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