Question

If y = x + tan x, show that  \[\cos^2 x\frac{d^2 y}{d x^2} - 2y + 2x = 0\] ?

Answer 1

Here,

\[y = x + \tan x\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = 1 + \sec^2 x\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = 2 \sec^2 x\tan x\]
\[\text { Dividing both sides by }\sec^2 x, \text { we get }\]
\[ \cos^2 x \frac{d^2 y}{d x^2} = 2\tan x\]
\[ \Rightarrow \cos^2 x \frac{d^2 y}{d x^2} = 2(y - x) \left[ \because y = x + \tan x \Rightarrow \tan x = y - x \right]\]
\[ \Rightarrow \cos^2 x \frac{d^2 y}{d x^2} - 2y + 2x = 0\]

Hence proved.