Question

If $y=x+tanx$, show that $co{s}^{2}x\frac{d^{2}y}{d{x}^2}-2y+2x=0$.

Answer 1

$y=x+\tan x$

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

$\frac{d^{2}y}{d{x}^2}=2\sec xsecx\tan x=2{\sec }^{2}xtanx$

Thus, ${\cos }^{2}x\frac{d^{2}y}{d{x}^2}-2y+2x=2{\sec }^{2}x\tan x{\cos }^{2}x-2x-2\tan x+2x=0$