Answer
See below
Work Step by Step
Given: $y_H(x)=c_1e^{-\int p(x)dx}$
a) Obtain: $y'_H(x)=c_1e^{-\int p(x)dx}\frac{d}{dx}(-\int p(x)dx)\\
=-c_1p(x)e^{-\int p(x)dx}$
We have: $y'_H(x)+p(x)y_H(x)\\=-c_1p(x)e^{-\int p(x)dx}+p(x)c_1e^{-\int p(x)dx}\\=0$
The integration factor is: $I(x)=e^{\int p(x)dx}$
Multiply equation by integrating factor:
$e^{\int p(x)dx}y'+ye^{\int p(x)dx}=0\\
\rightarrow \frac{d}{dx}(ye^{\int p(x)dx})=0\\
\rightarrow c_1=ye^{\int p(x)dx}\\
\rightarrow y_H(x)=c_1e^{-\int p(x)dx}$
b) Given: $y(x)=u(x)e^{-\int p(x)dx}$
then $y'+p(x)y\\=u'(x)e^{-\int p(x)dx}+u(x)e^{-\int p(x)dx}(-p(x))+p(x)u(x)e^{-\int p(x)dx}\\
=u'(x)e^{-\int p(x)dx}$
Let $q(x)=u'(x)e^{-\int p(x)dx}$
Thus, $q(x)=u'(x)e^{-\int p(x)dx}\\
\rightarrow u(x)=\int q(x)e^{\int p(x)dx}dx$
