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Solving this equation: $$y{\,^{iv}} + 5y\,''' - y\,'' + 8y\,' - 3y = 0$$ get a characteristic equation whose polynomial of 4th graders can not be factored by any known method, is not even factored by 2 2nd degree polynomials with pairs of complex roots. $$ {r} ^ {4} +5 \ {r} ^ {3} - {r} ^ {2} +8 \, r-3 = 0 $$ then: how would you find the solution of this ODE?

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    A degree 4 polynomial with real coefficients *must* either have a real root, or be equal to the product of two irreducible quadratics; your assertion that "it is not even factored by two 2nd degree polynomials with pairs of complex roots" is therefore necessarily false. Once you find the complex roots, $\rho_1,\ldots,\rho_4$, and the solutions are linear combinations of the complex exponentials $e^{\rho_it}$.2011-07-26
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    In fact, this polynomial *has* real roots: evaluating at $0$ you get $-3$, evaluating at $1$ you get $10$; by the intermediate value theorem, there must be a root between $0$ and $1$ (though, by the rational root theorem, it is irrational). In particular, there are at least two real roots. In any case, there are [formulas to solve a quartic by radicals](http://en.wikipedia.org/wiki/Quartic_equation#Solving_a_quartic_equation).2011-07-26
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    Just a small nitpick, put "$y^{iv}$" looks out of place with regular "$y'$"s. I prefer to write 4th degree or higher differential equations using a differential operator, i.e. "$D^4y + 5D^3y - D^2y + 8Dy - 3y = 0$"2011-07-26
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    @Arturo Magidin: Why does the rational root theorem imply an irrational root for this? Couldn't the root be `r = 1/3`? (It isn't).2011-07-26
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    @Christian: If $p(x)$ is a polynomial with integer coefficients, and $r/s$ is a rational root with $\gcd(r,s)=1$, then $r$ divides the constant term and $s$ divides the leading term. Since this polynomial is monic, rational roots would necessarily be integral. (Do you have the divisibility conditions on the rational root test reversed, perhaps?)2011-07-26
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    WolframAlpha will work out the pair of real and the pair of complex roots of the characteristic polynomial for you, both in numeric approximation and in symbolic form (by radicals).2011-07-26
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    @Arturo Magidin: Thanks. I had them reversed.2011-07-26
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    @mathsolomon: It might be worth noting that you can express the solutions in terms of the roots (even if you cannot write down the roots); but you *do* need to test to see if the polynomial has multiple roots or not. If it has no multiple roots, then the solutions are just linear combinations of $e^{\rho_i t}$, where $\rho_1,\rho_2,\rho_3,\rho_4$ are the roots; but if it *has* multiple roots, the expression changes a bit. Luckily, one can test to see if the polynomial has multiple roots without having to find the roots.2011-07-26

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There is in fact a thoroughly unpleasant closed form formula for the roots of a quartic. It was discovered by Ferrari, not the one of automobile fame, in the sixteenth century.

But the existence of a closed form formula is irrelevant. If you have a linear differential equation with constant coefficients, of any order, say for simplicity with no multiple roots, there is a simple expression for the solutions of the DE in terms of the roots of a certain polynomial.

Ultimately, we may end up having to approximate these roots. That is a familiar situation. Even when we have a closed form solution, to get numbers out we often need to approximate.

Added: Note that any polynomial with real coefficients can be factored as a product of linear and/or quadratic polynomials with real coefficients. There may not be a simple expression for these coefficients in terms of the coefficients of the original polynomial. But the coefficients of the factors can be found to any desired degree of accuracy.

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Wolfram Alpha reports four roots, two real and two complex. It declines to give expressions in terms of radicals, just numeric answers.

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    You can get the expressions in terms of radicals [here](http://www.wolframalpha.com/input/?i=r^4%2B5r^3-r^2%2B8r-3%3D0) by clicking the "Exact Forms" buttons. They're stupid complicated though.2011-07-26