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Euler–Cauchy equation

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The general second-order Euler–Cauchy equation is:

where a, b, and c can be any (real) constants.

Solving by the Method of Frobenius

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The leading coefficient function is 0 at ; so there is a singular point at . Normal form is:

where and .

Then and . These two functions, and , have no discontinuities; so the singular point is regular. Thus the method of Frobenius is applicable to this E.C.e. at .

Let

be a tentative Frobenius-series solution. Then

Substituting these into the E.C.e. yields

Since the L.H.S. is identically equal to 0, then each coefficient of the Frobenius series must equal 0:

When ,

Letting , then

is the indicial equation. Its roots are

If there exists some integer such that

then either or (since and are the only two solutions of the indicial equation); and there can be at most one such . For all other values of , the equation

is not satisfied, which means that for all such that and . (N.B.: If is not an integer then such an does not exist.)

or
or

The two right sides of the two logical disjunctions cannot be simultaneously true. Since is independent of , then a solution such as is a linear combination of linearly independent solutions.

Assuming that then this Wronskian is never zero, so the two terms of the solution are actually linearly independent solutions. (The case when will be dealt with further down from here.) The second term of is just “echoing” the other solution , so to speak. So the general solution is

The case of complex conjugate indicial roots

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If are a complex conjugate pair,

then

If is set to and is set to as well then

as may be derived through Euler's formula . If and then

These two solutions are linearly independent as can be verified by calculating their Wronskian:

which is non-zero almost everywhere. Thus

is a general solution.

The case of equal indicial roots

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If , let ;

is one solution. The other one may be determined through the method of undetermined coefficients. Let

where is a function of . Now derive:

The original E.C.e. is

Substitute and its derivatives into the E.C.e.:

Distribute:

Group by derivatives of :

Factor out powers of :

The expression within the square brackets is the indicial equation, for which is the root, so the third term in the sum of the L.H.S. vanishes.

For , , so divide by :

And here note well that , thus

Let , so that :

assuming that .

So and the general solution is

Solving another way

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Recalling the second order Euler–Cauchy equation:

Let . This is suggested by the solution corresponding to the indicial root , with instead of and instead of . So, letting , what happens when is derived with respect to instead of w.r.t. ?

where .


so


So

becomes

which is a “H.O.L.D.E.” (Homogeneous Ordinary Linear Differential Equation) with constant coefficients and characteristic equation

If its two roots are real and distinct then the H.O.L.D.E.w.c.c. has solution

where so the E.C.e. has solution


If the characteristic equation has two equal (and thus real) roots then the H.O.L.D.E.w.c.c. has solution

so the E.C.e. has solution


If the roots are a complex conjugate pair then the H.O.L.D.E.w.c.c. has solution

where, again, so the E.C.e. has solution


Higher-order cases

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This can be generalized:

To prove it, use mathematical induction. The base cases have already been proven. As inductive hypothesis, take this very formula that is to be proven. Divide it by and apply to (both sides of) it:

Thus the inductive step has been proven; the generalization is true:


This rule can be used to convert a higher-order E.C.e. into a H.O.L.D.E. with constant coefficients whose independent variable is z. Once the H.O.L.D.E.w.c.c. is solved, replace z with .

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