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?
How do I solve the ODE $y^{iv} + 5y'''- y''+ 8y' - 3y = 0 $?
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5A 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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1In 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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2Just 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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0@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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0@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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0WolframAlpha 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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0@Arturo Magidin: Thanks. I had them reversed. – 2011-07-26
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0@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
2 Answers
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.
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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1You 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