Question

If $y=x+\tan x$, prove that ${\cos }^{2}x\frac{d^{2}y}{d{x}^2}+2x=2y$

Answer 1

Given $y=x+tanx$

Differentiate with respect to $x$ we get

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

Again differentiate with respect to $x$ we get

$\frac{d^{2}y}{d{x}^2}=0+2\sec x(\sec x\cdot \tan x)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=2{\sec }^{2}x\cdot \tan x$

$\Rightarrow \frac{1}{se{c}^{2}x}\frac{d^{2}y}{d{x}^2}=2\tan x$

$\Rightarrow {\cos }^{2}x\frac{d^{2}y}{d{x}^2}=2(y-x)$

$\Rightarrow {\cos }^{2}x\frac{d^{2}y}{d{x}^2}=2y-2x$

$\Rightarrow {\cos }^{2}x\frac{d^{2}y}{d{x}^2}+2x=2y$

Hence proved.