Chapter 8.7, Problem 15P

If you solve (7.17) when $f(x)=0$ by assuming a solution $y=x^k,$ show that the quadratic equation for $k$ is the same as the auxiliary equation for the $z$ equation (7.20). Thus show (see Section 5) that if the two values of $k$ are equal, the second solution is not a power of $x$ but is $x^k$ ln $x$.

Also show that if $k$ is complex, say $k=a\pm bi,$ the solutions are $x^a\cos (b\ln x)$ and $x^a\sin (b\ln x)$ or other equivalent forms [see (5.16) to (5.18)].