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
B
1
C
2
D
3

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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