
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
$\displaystyle \frac{d^{3}y}{dx^{3}} + \frac{d^{2}y}{dx^{2}}  2y = 0$
I need to know what is the simplest way to solve these higher order DE
Please help(Headbang)

That's a linear DE with constant coefficients and its 'characteristic equation' is...
$\displaystyle d^{3} + d^{2}  2 = 0$ (1)
But ...
$\displaystyle d^{3} + d^{2}  2 = (d^{2} + 2d +2)\cdot (d1)$ (2)
... so that...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$

Thx Chi
how did you get that, I how do see that? is there any special method you used

In this case it is immediate 'discovering' that $\displaystyle d=1$ makes null the polynomial $\displaystyle d^{3} + d^{2} 2$, so that $\displaystyle d1$ 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
$\displaystyle \chi$ $\displaystyle \sigma$