For the Euler-Cauchy equation with complex roots m = -1 ± i, which of the following is a correct form of the general real solution?

• A. y = x [ C1 cos(ln x) + C2 sin(ln x) ]
• B. y = x^{-2} [ C1 cos(ln x) + C2 sin(ln x) ]
• C. y = x^{-1} [ C1 cos(ln x) + C2 sin(ln x) ]
• D. y = x^{-1} [ C1 cosh(ln x) + C2 sinh(ln x) ]

Answer: (C)

When solving an Euler-Cauchy equation, you try a power solution y = x^m. Substituting in gives a quadratic for m, and if the roots are complex α ± iβ, the real solutions come from combining x^α cos(β ln x) and x^α sin(β ln x). This happens because x^{α+iβ} = x^α e^{iβ ln x} = x^α [cos(β ln x) + i sin(β ln x)], and taking real and imaginary parts yields two independent real solutions.

Here the roots are -1 ± i, so α = -1 and β = 1. The real general solution is y = x^{-1} [C1 cos(ln x) + C2 sin(ln x)], valid for x > 0. The other forms would require a different α or replace the trig functions with hyperbolic ones, which does not match these complex roots.