It is given that, y = 3cos(log x) + 4sin(log x)
Then,
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = 3 × $\frac{\mathit{d}}{\mathit{d}\mathit{x}}$ [cos log x] + 4 × $\frac{\mathit{d}}{\mathit{d}\mathit{x}}$ [sin log x]
= 3 × $-\mathit{s}\mathit{i}\mathit{n}\mathit{l}\mathit{o}\mathit{g}\mathit{x}\times \frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathit{l}\mathit{o}\mathit{g}\mathit{x}]$ + 4 × $\left[\mathit{c}\mathit{o}\mathit{s}\mathit{l}\mathit{o}\mathit{g}\mathit{x}\times \frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right]$
= $\frac{-3\mathit{s}\mathit{i}\mathit{n}\mathit{l}\mathit{o}\mathit{g}\mathit{x}}{\mathit{x}}$ + $\frac{4\mathit{c}\mathit{o}\mathit{s}\mathit{l}\mathit{o}\mathit{g}\mathit{x}}{\mathit{x}}$ = $\frac{4\mathit{c}\mathit{o}\mathit{s}\mathit{l}\mathit{o}\mathit{g}\mathit{x}-3\mathit{s}\mathit{i}\mathit{n}\mathit{l}\mathit{o}\mathit{g}\mathit{x}}{\mathit{x}}$
$\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}$ = $\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left(\frac{4\mathit{c}\mathit{o}\mathit{s}\left(\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right)-3\mathit{s}\mathit{i}\mathit{n}\left(\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right)}{\mathit{x}}\right)$
=
=
=
=
∴ ${\mathit{x}}^2\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}$ + x $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ + y = ${\mathit{x}}^2\left(\frac{-\mathit{s}\mathit{i}\mathit{n}\left(\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right)-7\mathit{c}\mathit{o}\mathit{s}\left(\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right)}{{\mathit{x}}^2}\right)$ + x $\left(\frac{4\mathit{c}\mathit{o}\mathit{s}\left(\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right)-3\mathit{s}\mathit{i}\mathit{n}\left(\mathit{l}\mathit{o}\mathit{g}\mathit{x}\right)}{\mathit{x}}\right)$ + 3 cos (log x) + 4 sin (log x)
= - sin (log x) - 7 cos (log x) + 4 cos (log x) - 3 sin (log x) + 3 cos (log x) + 4 sin (log x)
= 0
Hence proved.