If y = x + tan x, show that cos² x (d² y)/(dx²) - 2y + 2x = 0.

If $y = x + \tan x$, show that $\cos^2 x \frac{d^2 y}{dx^2} - 2y + 2x = 0$.

Given: $y = x + \tan x$

To show: $\cos^2 x \frac{d^2 y}{dx^2} - 2y + 2x = 0$

Answer

With $y=x+\tan x$: $\frac{dy}{dx}=1+\sec^2x$, $\frac{d^2y}{dx^2}=2\sec^2x\tan x$, so $\cos^2x\frac{d^2y}{dx^2}=2\tan x$ and $-2y+2x=-2\tan x$; the sum is $0$.

1. $y=x+\tan x$.
2. Differentiate: $\frac{dy}{dx}=1+\sec^2x$.
3. Differentiate again: $\frac{d^2y}{dx^2}=2\sec^2x\tan x$.
4. Multiply by $\cos^2x$: $\cos^2x\frac{d^2y}{dx^2}=2\sec^2x\cos^2x\tan x=2\tan x$ (as $\sec x\cos x=1$).
5. Substitute $y=x+\tan x$: $-2y+2x=-2x-2\tan x+2x=-2\tan x$.
6. Add: $\cos^2x\frac{d^2y}{dx^2}-2y+2x=2\tan x-2\tan x=0$.