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I found this characteristic equation $r(r-1)^2=1$ in my homework Euler differential equation problem (page 669, again the book), $x^{3} y'''+xy'-y=0$, when I misread the problem as

$$x^{3} y'''+xy''-y=0$$

so this beast has characteristic equation that looks nightmarish at least here in WA. I am not sure whether I am missing some smart substitution to proceed, how would you solve the polynomial?

Please, keep the focus on the polynomial. Thanks for pointing out the err but I would like to know how to solve this kind of polynomials.

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3 Answers

up vote 4 down vote accepted

You can make the substitution $$ z= \displaystyle \frac{1}{r-1} $$ and obtain the equation:

$$z^3 - z - 1 = 0$$

This can be solved cleverly using the substitution $\displaystyle z = \frac{2 \cos \theta}{\sqrt{3}}$ (note: $\displaystyle \theta$ is allowed to be complex) giving rise to the equation

$$ \cos 3 \theta = \frac{3 \sqrt{3}}{2}$$

Thus we get the answer in terms of the complex function $\displaystyle \arccos$ (whose possible arguments range over all the complex numbers)

See here for an explanation of the method: Trigonometric and Hyperbolic Substitution to solve the cubic.

Of course, as Sasha points out, you probably are trying to solve the wrong equation in the first place.

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Because the ODE you are solving is scale invariant, i.e. if $y(x)$ solves it, so does $y(\lambda x)$ for arbitrary $\lambda \not=0 $, we seek solution as $y(x) = x^r$. This leads to the characteristic equation: $$ 0 = r(r-1)(r-2) + r - 1 = r^3 - 3 r^2 + 3r -1 = (r-1)^3 $$ Because the root $r=1$ has triple multiplicity, the solution is logarithmic: $$ y(x) = x( c_1 + c_2 \ln(x) + c_3 \ln^2(x) ) $$

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...this makes the chapter X.6. in the book much clearer, thanks. –  hhh Feb 24 '12 at 21:12
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You could try using the cubic formula.

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Since the polynomial has coefficients in $\mathbb{Z}$, you could try the rational root theorem. This procedure will fail in this case, as the polynomial has no rational roots. For polynomials of degree two to four, there exist formulas for roots in terms the coefficients (as mentioned above). Otherwise, you could try a numerical method. If this doesn't suit you, read about the Abel-Ruffini theorem. –  dls Feb 24 '12 at 20:47
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