Answer
$$ y= \frac{1}{2}\left(\sin x-\cos x\right)+ \frac{3}{2}e^{-x}$$
Work Step by Step
Given$$y^{\prime}+y=\sin x, \quad y(0)=1$$
This is a linear equation with $p(x) = 1,\ \ q(x) =\sin x$, so
\begin{align*}
\mu(x)&=e^{\int p(x)dx}\\
&=e^{\int dx}\\
&=e^x
\end{align*}
Then
\begin{align*}
y\mu(x) &=\int \mu(x)q(x)dx\\
e^{x}y &=\int e^x\sin xdx
\end{align*}
Use
$$\int e^{a u} \sin b u d u=\frac{e^{a u}}{a^{2}+b^{2}}(a \sin b u-b \cos b u)+C $$
Then
$$y= \frac{1}{2}\left(\sin x-\cos x\right)+ Ce^{-x} $$
Since $y(0)=1 $, then $C= \frac{3}{2}$, and hence
$$ y= \frac{1}{2}\left(\sin x-\cos x\right)+ \frac{3}{2}e^{-x}$$
