PX153 - E1 - recap and introduction

a_{2} (x) \frac{d^{2} y}{d x^{2}} + a_{1} (x) \frac{d y}{d x} + a_{0} (x) y (x) = f (x)

GS will have 2 constants of integration, so, we will need 2 boundary conditions
two steps:
- step 1: find the GS to the complementary equation - the homogenous version with $f (x) = 0$
  - this gives $y_{c} (x)$ the complementary function:
  $a_{2} (x) \frac{d^{2} y_{c}}{d x^{2}} + a_{1} (x) \frac{d y_{c}}{d x} + a_{0} (x) y_{c} = 0 $ $ w h i c h w i l l h a v e 2 c o n s t a n t s o f i n t e g r a t i o n$
$a_{2} (x) \frac{d^{2} y_{p}}{d x^{2}} + a_{1} (x) \frac{d y_{p}}{d x} + a_{0} (x) y_{p} = f (x) $ $ w i t h n o a r b i t r a r y c o n s t a n t s$

\begin{matrix} a_{2} (x) \frac{d^{2} y}{d x^{2}} + a_{1} (x) \frac{d y}{d x} + a_{0} (x) y (x) = (a_{2} (x) \frac{d^{2} y_{c}}{d x^{2}} + a_{1} (x) \frac{d y_{c}}{d x} + a_{0} (x) y_{c}) \\ + (a_{2} (x) \frac{d^{2} y_{p}}{d x^{2}} + a_{1} (x) \frac{d y_{p}}{d x} + a_{0} (x) y_{p}) = 0 + f (x) \end{matrix}