Answer
See below
Work Step by Step
Given: $y''-8y'+17y=e^{4x}\csc x$
In this case the substitution $y(x) = x^r$ yields the indicial equation
$r^2-8r+17=0$
Solve the equation
$r_1=4+i\\
r_2=4-i$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=e^{4x}\sin x\\
y_2(x)=e^{4x}\cos x$
so that the particular solution is
$y(x)=u_1e^{4x}\sin x+u_2e^{4x}\cos x$
We first compute the appropriate derivatives
$0=u_1'(x)e^{4x}\sin x+u_2'(x)e^{4x}\cos x\\
e^{4x}\csc x=u_1'(x)(e^{4x}\sin x)'+u_2'(x)(e^{4x}\cos x)'$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$u_1'=\cot x\\
u_1=\ln (\sin x)$
and $u_2'=-1\\
u_2=-x$
Hence, the general solution is:
$y(x)=e^{4x}\sin\ln (\sin x)-e^{4x}x\cos x$
