Higher-Order DE factoring

Hi,

Can somebody please help me to factor the following DE?

$$2\frac{d^5y}{dx^5} -7\frac{d^4y}{dx^4} + 12\frac{d^3y}{dx^3} + 8\frac{d^2y}{dx^2} = 0$$

The auxiliary equation of above DE is

$$2m^5 - 7m^4 + 12m^3 + 8m^2 = 0$$

$$m^2(2m^3-7m^2 + 12m + 8) = 0$$

The equation of $$2m^3-7m^2+12m+8$$ is cannot be factored in any form, at least after I try for several times.

The equation of $$2m^3-7m^2+12m+8$$ is cannot be factored in any form, at least after I try for several times.

That polynomial can certainly be factorized. You just need to look a little harder, or try some basic root finding

Hi Scottie,

Since the degree of polynomial is > 2 then to find the factor is “by division”.

I’ve try that way but there is no solution.

Do you mean that “basic root finding” is$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$?

If so, since it’s only can be used for 2-degree polynomial, how to use that formula in 3 degree of polynomial equation?

Thank you very much

epenguin
Homework Helper
Gold Member
Just plot
$$(2m^3-7m^2 + 12m + 8)$$
with a plotting hand calculator or computer plotting routine or by hand calculation. You will see a root, then check you have an exact one.

Note you have already factored the equations somewhat; that gives you some solution already which will later enter into the general solution.

HallsofIvy
The "rational root theorem" says that the only possible rational roots must have denominator divisible by 2 (coefficient of x3) and numerator a factor of 8. That is, the only possible rational roots are $\pm 1/2$, $\pm$1, $\pm$2, $\pm$4, or $\pm$8. Since you can get reasonable factoring only with rational roots, try those and see if any are roots.
Graphing 2m3-7m2+12m+8 reveals that $$\frac{-1}{2}$$ is a root.
Using polynomial long division by (m+$$\frac{1}{2}$$) yields this:
(m+$$\frac{1}{2}$$)(2m2-8m+16)=2m3-7m2+12m+8