Question

If $x=\tan \left(\frac{1}{a}logy\right)$, prove that $(1+x^2)\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}-a0$.

Answer 1

Take ${\tan }^{-1}$ on both sides,

$\Rightarrow {\tan }^{-1}x=\frac{1}{a}\log y$

Differentiate w.r.t x,

$\frac{1}{x^2+1}=\frac{1}{ay}\frac{dy}{dx}$.....$\left(1\right)$

Again differentiate w.r.t $x(\because$ we need second order differential equation$)$

$-\frac{2x}{(1+x^2{)}^2}=-\frac{1}{a{y}^2}\frac{dy}{dx}+\frac{1}{ay}\frac{d^{2}y}{d{x}^2}$

$\frac{1}{ay}\frac{d^{2}y}{d{x}^2}+\frac{2x}{(1+x^2{)}^2}-\frac{1}{a{y}^2}\frac{dy}{dx}=0$

By $\left(1\right)$, substitute value of $\frac{1}{x^2+1}$

$\frac{1}{ay}\frac{d^{2}y}{d{x}^2}+\frac{2x}{(1+x^2)}\times \frac{1}{ay}\frac{dy}{dx}-\frac{1}{a{y}^2}\frac{dy}{dx}=0$

$\frac{d^{2}y}{d{x}^2}+\frac{2x}{(1+x^2)}\times \frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=0$

$(1+x^2)\frac{d^{2}y}{d{x}^2}+2x\times \frac{dy}{dx}-\frac{1+x^2}{y}\frac{dy}{dx}=0$

Now from $\left(1\right)$ substitute $\frac{1+x^2}{y}=a\frac{dx}{dy}$

$(1+x^2)\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}-a\frac{dx}{dy}\frac{dy}{dx}=0$

$(1+x^2)\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}-a=0$

Hence proved.