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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calculus
complex-analysis
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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