Find a homogeneous second-order Cauchy-Euler equation with real coefficients if the given number is a root of its auxiliary equation. mi= i
C1cos(ln(x)) + C2sin(ln(x))

I'm going to assume that you mean to say that i = √(-1) is a root of the auxiliary equation. That is, if the Cauchy-Euler DE is

x ²y'' + axy' + by = 0

then the auxiliary equation obtained by substituting y = xᵐ is

x ² (m (m - 1) xᵐ ⁻ ²) + ax (m xᵐ ⁻ ¹) + bxᵐ = 0

which reduces to

m (m - 1) + am + b = 0

or

m ² + (a - 1) m + b = 0

By the fundamental theorem of algebra, we can write the quadratic in terms of its roots r₁ and r₂,

(m - r₁) (m - r₂) = 0

Given that one root is the imaginary unit i, and the coefficients of the aux. equation are real, it follows that the other root is -i, because complex roots must occur with their conjugates. So we have as our aux. equation,

(m - i ) (m + i ) = 0

or

m ² + 1 = 0

Then a - 1 = 0 and b = 1, so that the given root and general solution correspond to the DE,

x ²y'' + xy' + y = 0

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation: