Question

$x=\tan \left(\frac{1}{a}\log y\right)prove:\left(1+x^2\right)y_2+\left(2x-a\right)y_1=0$

Answer 1

We have,

$x=\tan \left(\frac{1}{a}\log y\right)$

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

On differentiating w.r.t $x$, we get

$\frac{1}{1+x^2}=\frac{1}{ay}\frac{dy}{dx}$

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

Again, differentiating w.r.t $x$, we get

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

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

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

Put $\frac{dy}{dx}=y_1,\frac{d{y}^2}{d{x}^2}=y_2$

Therefore,

$(1+x^2)y_2+(2x-a)y_1=0$

Hence, proved.