Higher-Order DE Factoring: Can You Help Me Factor This Differential Equation?

[optics.tech]

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.

Thanks in advance

 

[scottie_000]

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

 

[optics.tech]

Hi Scottie,

Thanks in advance.

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]

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.

 

[JeffreyBlake]

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

See if you can finish from here.