Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 3

$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$

Hint:

You have to show

$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$

Given:

If y=x+tan x show that

$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$

Solution:

Let y=x+tan x

$y=x+tan x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (\frac{d\: tan\: x}{dx}=sec^{2}x,\: \frac{\mathrm{d} }{\mathrm{d} x}x=1)$

\begin{aligned} &\frac{d y}{d x}=1+\sec ^{2} x \\ &\frac{d^{2} y}{d x^{2}}=2 \sec x(\sec x \tan x) \quad\left(\frac{d}{d x} \sec ^{2} x=2 \sec x(\sec x \tan x)\right. \\ &\frac{d^{2} y}{d x^{2}}=2 \sec ^{2} x \tan x \quad \quad\left(\sec ^{2} x=\frac{1}{\cos ^{2} x}, \tan x=\frac{\sin x}{\cos x}\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{2 \sin x}{\cos ^{3} x} \end{aligned}

According to question

$\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0$

substituting these values we get

\begin{aligned} &\left(\frac{2 \sin x}{\cos x}\right)-2(x+\tan x)+2 x \\ &\frac{2 \sin x}{\cos x}-2 x-2 \tan x+2 x \\ &2 \tan x-2 x-2 \tan x+2 x=0 \end{aligned}

Hence proved