# Easiest way to solve

• Sep 2nd 2009, 11:43 PM
khanim
Easiest way to solve
Hi(Cool)

The section that I am currently busy with is the D operator, I am learning how to find the aux equation and complimentary functions

Its basic maths but unfortunately I am struggling with the higher DE

$\frac{d^{3}y}{dx^{3}} + \frac{d^{2}y}{dx^{2}} - 2y = 0$

I need to know what is the simplest way to solve these higher order DE
• Sep 2nd 2009, 11:55 PM
chisigma
That's a linear DE with constant coefficients and its 'characteristic equation' is...

$d^{3} + d^{2} - 2 = 0$ (1)

But ...

$d^{3} + d^{2} - 2 = (d^{2} + 2d +2)\cdot (d-1)$ (2)

... so that...

Kind regards

$\chi$ $\sigma$
• Sep 3rd 2009, 12:34 AM
khanim
Thx Chi
how did you get that, I how do see that? is there any special method you used
• Sep 3rd 2009, 01:16 AM
chisigma
In this case it is immediate 'discovering' that $d=1$ makes null the polynomial $d^{3} + d^{2} -2$, so that $d-1$ is a factor of that polynomial. In general we aren't so 'lucky' and in such a case we have to 'attack' a third order algebraic equation, that the Italian mathematicians Scipione del Ferro and Niccolò Fontana first successfully resolved between fifteenth and sixteenth century (Rofl) ...

Kind regards

$\chi$ $\sigma$