Question

If y =sin (log x), then ${\displaystyle x^2\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}+x\frac{\text{dy}}{\text{dx}}+\text{y}}$ is equal to ______.

Options

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Answer 1

If y =sin (log x), then ${\displaystyle x^2\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}+x\frac{\text{dy}}{\text{dx}}+\text{y}}$ is equal to 0.

Explanation:

y = sin (log x)     ...(i)

${\displaystyle \therefore \frac{\text{dy}}{\text{dx}}=\cos \left(\log x\right)\cdot \frac{1}{x}}$

${\displaystyle \Rightarrow x\frac{\text{dy}}{\text{dx}}=\cos \left(\log x\right)}$

Differentiating both sides w.r.t. x, we get

${\displaystyle x\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}+\frac{\text{dy}}{\text{dx}}\cdot 1=-\sin \left(\log x\right)\cdot \frac{1}{x}}$

${\displaystyle \Rightarrow {\text{x}}^2\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}+x\frac{\text{dy}}{\text{dx}}}$ = - y    ....[From (i)]

${\displaystyle \Rightarrow {\text{x}}^2\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}+x\frac{\text{dy}}{\text{dx}}}$+ y = 0