## Thomas' Calculus 13th Edition

$$y = \ln |\sec x|+x$$
Given $$\frac{d^2y}{dx^2}=\sec^2 x ,\ \ \ \ \ \ y(0)=0,\ \ \ \ y'(0)=1$$ Then \begin{align*} y'&= \int \sec^2 x dx\\ &= \tan x +c \end{align*} At $x=0$ , $y'= 1$, then $c=1$ and $$y' = \tan x+1$$ Hence \begin{align*} y&= \int (\tan x+1) dx\\ &= \ln |\sec x|+x+c \end{align*} At $x=0$, $y=0$, then $c=0$ and $$y = \ln |\sec x|+x$$