Answer
$$ y(x)=1 $$
Work Step by Step
Given$$y^{\prime}+(\sec t) y=\sec t, \quad y\left(\frac{\pi}{4}\right)=1$$
This is a linear equation with $p(t) =\sec t\ \ q(t) =\sec t$, so
\begin{align*}
\mu(t)&=e^{\int p(t)dt}\\
&=e^{\int \sec t dt}\\
&=e^{\ln |\sec t+\tan t |}\\
&=\sec t+\tan t
\end{align*}
Then
\begin{align*}
y\mu(t) &=\int \mu(t)q(t)dt\\
(\sec t+\tan t ) y &=\int (\sec^2 t+\tan t\sec t ) dt\\
&= \sec t+\tan t+C
\end{align*}
Then $$y=1+\frac{C}{\sec t+\tan t}$$
Since $y(\pi/4)=1 $, then $C=0$, and hence
$$ y(x)=1 $$
