Answer
$$ y = \ln |\sec x|+x$$
Work Step by Step
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
$$ y = \ln |\sec x|+x$$
