Answer
See below
Work Step by Step
Given $y''+y=\csc x$
with $y_1=\sin x$
We first compute the appropriate derivatives
$y'=\sin x(u'+u)\\
y''=\sin x(u''+2u'-u)$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$\sin x(u''+2u'-u)+\sin x(u'+u)=\csc x$
which simplifies to
$\sin xu''+2u'\cos x=\csc x$
or equivalently
$w'+2\cos x w=\csc x$
where $w = u′$
An integrating factor is
$I=e^{2\int \cot xdx}=2e^{2\ln (\sin x)}=\sin^2x$
So that
$\frac{d}{dx}(\sin^2 x w)=\sin^2x\csc^2x$
Integrating both sides with respect to x yields
$\sin^2x w=x+C_1$
where $c_1$ is a constant. Thus
$u'(x)=x\csc^2x+C_1\csc^2x$
which can be integrated directly to obtain
$u(x)=-x\cot x+\ln (\sin x)-C_1\cot x+C_2$
The general solution is
$y(x)=-C_1\cos x+C_2\sin x+\sin x \ln (\sin x)--x\cos x$
