Which of the following is the general solution to the Euler-Cauchy equation x^2 y'' + 3 x y' + 2 y = 0?

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Multiple Choice

Which of the following is the general solution to the Euler-Cauchy equation x^2 y'' + 3 x y' + 2 y = 0?

When solving an Euler-Cuchy (Euler) equation, try a power solution y = x^r. Substituting gives r(r-1) x^r + 3 r x^r + 2 x^r = 0, so r^2 + 2r + 2 = 0. The roots are r = -1 ± i. For complex roots α ± iβ, the real general solution is y = x^{α} [C1 cos(β ln x) + C2 sin(β ln x)]. Here α = -1 and β = 1, yielding y = x^{-1} [C1 cos(ln x) + C2 sin(ln x)]. This is the standard real-valued form. The exponential form y = C1 x^{-1+i} + C2 x^{-1-i} is algebraically equivalent (if C1, C2 can be complex), but the real general solution uses the sine and cosine form shown.

When solving an Euler-Cuchy (Euler) equation, try a power solution y = x^r. Substituting gives r(r-1) x^r + 3 r x^r + 2 x^r = 0, so r^2 + 2r + 2 = 0. The roots are r = -1 ± i. For complex roots α ± iβ, the real general solution is y = x^{α} [C1 cos(β ln x) + C2 sin(β ln x)]. Here α = -1 and β = 1, yielding y = x^{-1} [C1 cos(ln x) + C2 sin(ln x)]. This is the standard real-valued form. The exponential form y = C1 x^{-1+i} + C2 x^{-1-i} is algebraically equivalent (if C1, C2 can be complex), but the real general solution uses the sine and cosine form shown.

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