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katrin [286]
3 years ago
14

Find a homogeneous second-order Cauchy-Euler equation with real coefficients if the given number is a root of its auxiliary equa

tion.
mi= i
C1cos(ln(x)) + C2sin(ln(x))
Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0

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

<em>x</em> ²<em>y''</em> + <em>axy'</em> + <em>by</em> = 0

then the auxiliary equation obtained by substituting <em>y</em> = <em>xᵐ</em> is

<em>x</em> ² (<em>m</em> (<em>m</em> - 1) <em>xᵐ </em>⁻<em> </em>²) + <em>ax</em> (<em>m</em> <em>xᵐ </em>⁻<em> </em>¹) + <em>bxᵐ</em> = 0

which reduces to

<em>m</em> (<em>m</em> - 1) + <em>am</em> + <em>b </em>= 0

or

<em>m</em> ² + (<em>a</em> - 1) <em>m</em> + <em>b</em> = 0

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

(<em>m</em> - <em>r₁</em>) (<em>m</em> - <em>r₂</em>) = 0

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

(<em>m</em> - <em>i</em> ) (<em>m</em> + <em>i </em>) = 0

or

<em>m</em> ² + 1 = 0

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

<em>x</em> ²<em>y''</em> + <em>xy'</em> + <em>y</em> = 0

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