PX153 - E2 - method of undetermined coefficients

just a posh way of saying PX153 - E1 - recap and introduction
method of finding the particular integral
- essentially "guess" a trial solution based on the form of $f (x)$
- substitute it into the ODE and find the undetermined coefficients

$f (x)$	trial solution
polynomial of degree $n$ , eg: $x^{3} + 2 x + 1$	polynomial of degree $n$ , eg: $y_{p} = α x^{3} + β x^{2} + γ x + δ$
$\cos q x o r \sin q x$	$y_{p} = α \cos q x + β \sin q x$
$p e^{q} x$	$y_{p} (x) = α e^{q x}$

similarly, in the case of a repeated root in the auxiliary equation, we can have a problem if the usual trial function is already in the complementary function. as with the repeated root, our solution is to multiply by the independent variable.
- eg: find the general solution to:
$y^{″} - y^{'} - 6 y = e^{3 x}$ - complementary function from before
$$y_{c}(x) = Ce^{3x}+De^{2x}$$
- trial function for $e^{3 x}$ will not work
- instead, we try $y_{p} (x) = α x e^{3 x}$
$$y_{p}' = \alpha e^{3x}+3\alpha xe^{3x}$$
$$y_{p}'' = 6 \alpha e^{3x}+ 9\alpha xe^{3x}$$
$$(6 \alpha e^{3x}+ 9\alpha xe^{3x})-(\alpha e^{3x}+ 3\alpha xe^{3x})-6\alpha x e^{3x}=e^{3x}$$
- terms in $x e^{3 x} : 9 α - 3 α - 6 α = 0$
- terms in $x^{0} e^{3 x} : 6 α - α = 1 ⟹ α = \frac{1}{5}$
$$\therefore G.S. : y(x) = y_{c}+ y_{p} = Ce^{3x}+De^{2x} + \frac{1}{5} e^{3x}$$
if $f (x)$ contains a sum of functions, then our trial function will also be the sum of the functions
- eg: for $a y^{″} + b y^{'} + c y = 3 \sin (2 x) + 1 + x^{2}$
  - try: $α \sin (2 x) + β \cos (2 x) + (γ + δ x + ϵ x^{2})$
note: the trial functions we gave assumed an ODE of the form $a y^{″} + b y^{'} + c y = f (x)$
- if this is not the case, think more carefully
- eg: solve $y^{″} - 2 y^{'} = - 5 \cos (x) - 2$
  - no term in $y$
  - first, solve complementary equation:
    - auxiliary equation: $l^{2} - 2 l = 0$
      $l_{1} = 0, l_{2} = 2$
    $y_{c} = A + B e^{2 x}$
  - particular solution: $y_{p} (x) = C \cos (x) + D \sin (x) + E x$
    - sub into ODE: $- C \cos (x) - D \sin (x) + 2 C \sin (x) - 2 D \cos (x) - 2 E = - 5 \cos (x) - 2$
    - constant term: $- 2 E = - 2 ⟹ E = 1$
    - $\cos (x)$ term: $- C - 2 D = - 5$
    - $\sin (x)$ term: $- D + 2 C = 0 ⟹ D = 2 C$
      $∴ C = 1, D = 2$
    - hence,
    $∴ G . S . : y (x) = A + B e^{2 x} + \cos (x) + 2 \sin (x) + x$