Question

$\mathrm{If}x=\tan \left(\frac{1}{a}\log y\right),\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\left(1+x^2\right)\frac{d^{2}y}{{\mathrm{dx}}^2}+\left(2x-a\right)\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{is}$

Answer 1

We have,
$x=\tan \left(\frac{1}{a}\log y\right)\Rightarrow {\tan }^{-1}x=\frac{1}{a}\log y\Rightarrow a{\tan }^{-1}x=\log y\mathrm{Differentiating}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \frac{a}{1+x^2}=\frac{1}{y}\frac{\mathrm{dy}}{\mathrm{dx}}\Rightarrow \left(1+x^2\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ay}\mathrm{Differentiating}\mathrm{again}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \left(1+x^2\right)\frac{d^{2}y}{{\mathrm{dx}}^2}+2x\frac{\mathrm{dy}}{\mathrm{dx}}=a\frac{\mathrm{dy}}{\mathrm{dx}}\Rightarrow \left(1+x^2\right)\frac{d^{2}y}{{\mathrm{dx}}^2}+\left(2x-a\right)\frac{\mathrm{dy}}{\mathrm{dx}}=0$