Higher Order DE

homogeneous Equations¶

Definition¶

a_{n} (x) \frac{d^{n} y}{d x^{n}} + a_{n - 1} (x) \frac{d^{n - 1} y}{d x^{n - 1}} + \dots + a_{1} (x) \frac{d y}{d x} + a_{0} (x) y = g (x)

or be written as

a_{n} (x) D^{n} + a_{n - 1} (x) D^{n - 1} + \dots + a_{1} (x) D + a_{0} (x) y = g (x)

$g (x) = 0 \to$ homogeneous
$g (x) \neq 0 \to$ non-homogeneous

General Solution (Complementary Function)¶

Theorem 4.1.5

For an n^th order homogeneous liner DE $L (y) = 0$ , if
1. $y_{1} (t), y_{2} (t), \dots, y_{n} (t)$ are the solution to $L (y) = 0$
2. $y_{1} (t), y_{2} (t), \dots, y_{n} (t)$ are linearly independent
3. Determinate whether they are liner independent : Wronskian.
Expression

Any solution of the homogeneous liner DE can be expressed as

$y = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n}$
Any n^th order homogeneous linear DE has n linearly independent solutions.
fundamental set of solutions

$y_{1} (t), y_{2} (t), \dots, y_{n} (t)$
general solution (aka complementary function)

y = c_{1} y_{1} + c_{2} y_{2} + \dots + c_{n} y_{n}

2nd with an known solution¶

2^nd order linear homogeneous DE with an known solution

a_{2} (x) y^{″} + a_{1} (x) y^{'} + a_{0} (x) y = 0

By virtue of solution below, we can get another solution and therefore get complementary function.

Conditions¶

second order
linear
homogeneous
one of the nontrivial solution $y_{1} (x)$ has been known

#### Standard Form

y^{″} + P (x) y^{'} + Q (x) y = 0

Solution¶

(直接背！！！)

y_{2} (x) = y_{1} (x) \int \frac{e^{- \int P (x) d x}}{y_{1}^{2} (x)} d x

see proof. (TODO)

Linear DE with Constant Coefficients¶

Homogeneous linear DE with constant coefficients

a_{n} y^{(n)} (x) + a_{n - 1} y^{(n - 1)} (x) + \dots + a_{1} y^{'} (x) + a_{0} y = 0

Condition¶

homogeneous
linear
constant coefficients

kernel concept¶

Suppose the solutions has the form of $e^{m x}$

Auxiliary Function¶

change $y^{(n)}$ into $m^{n}$

a_{n} m^{n} + a_{n - 1} m^{n - 1} + \dots + a_{1} m + a_{0} = 0

solve $m$ .

Solution to 2nd Order¶

\begin{matrix} a_{2} y^{″} (x) + a_{1} y^{'} (x) + a_{0} y (x) = 0 \\ ↓ \\ a_{2} m^{2} + a_{1} m + a_{0} = 0 \end{matrix}

Case 1 : $m_{1} \neq m_{2}, m_{1}, m_{2} \in ℝ$ , (D > 0)

$m_{1}, m_{2} = \frac{- a_{1} \pm \sqrt{a_{1}^{2} - 4 a_{2} a_{0}}}{2 a_{2}}$

thus

$y = c_{1} e^{m_{1} x} + c_{2} e^{m_{2} x}$

if

$m_{1}, m_{2} = α \pm β$

we can also write

$y = e^{α x} (c_{1} \cosh β x + c_{2} \sinh β x)$

Case 2 $m_{1} = m_{2}$ , (D = 0)

$y_{1} = e^{m_{1} x}$

$y_{2}$ can be find by the method mentioned above.

then we found that

$y_{2} = x e^{m_{1} x}$
Case 3 $m_{1} \neq m_{2}$ , $m_{1}$ and $m_{2}$ are conjugate(共軛) and complex, (D < 0)

$\begin{matrix} m_{1} = α + i β m_{2} = α - i β \\ α = - \frac{a_{1}}{2 a_{2}}, β = \frac{\sqrt{4 a_{2} a_{0} - a_{1}^{2}}}{2 a_{2}} \end{matrix}$

thus

$\begin{matrix} y = C_{1} e^{α x + i β x} + C_{2} e^{α x - i β x} \end{matrix}$

another form : proof

$y = e^{α x} (c_{1} \cos β x + c_{2} \sin β x)$

Solution to Higher Order¶

n^th order ODE

$p, q \in [1, n], and p \neq q$

Case 1 : $m_{p} \neq m_{q}$ 都是獨立解
Case 2 : 有重根 (在 $m_{p}$ 處重根 k 個)

solution : $e^{m_{p} x}, x \cdot e^{m_{p} x}, x^{2} \cdot e^{m_{p} x}, \dots, x^{k - 1} \cdot e^{m_{p} x}$
Case 3 : 有k對複數解

solutions :

$\begin{array}{r} e^{α x} \cos (β x), x e^{α x} \cos (β x), x^{2} e^{α x} \cos (β x), \dots, x^{k - 1} e^{α x} \cos (β x) \\ e^{α x} \sin (β x), x e^{α x} \sin (β x), x^{2} e^{α x} \sin (β x), \dots, x^{k - 1} e^{α x} \sin (β x) \end{array}$

Cauchy-Euler Equation¶

Cauchy-Euler Equation is homogeneous linear DE in the form below

a_{n} x^{n} y^{(n)} (x) + a_{n - 1} x^{n - 1} y^{(n - 1)} (x) + \dots + a_{1} x y^{'} (x) + a_{0} y (x) = g (x)

Kernel Concept¶

guess the solution has the form $y (x) = x^{m}$

then we can change $x^{k} \frac{d^{k}}{d x^{k}}$ to $\frac{m!}{(m - k)!}$

Solution to 2nd Order¶

a_{2} x^{2} y^{″} (x) + a_{1} x y^{'} (x) + a_{0} y = 0

auxiliary function :

$\begin{aligned} a_{2} m (m - 1) & + a_{1} m & + a_{0} & = 0 \\ a_{2} m^{2} & + (a_{1} - a_{2}) m & + a_{0} & = 0 \end{aligned}$

than solve roots with

\begin{array}{r} m = \frac{- b \pm \sqrt{D}}{2 a} \end{array}

Case 1 : ( $m_{1} \neq m_{2}, m_{1}, m_{2} \in ℝ$ )

$y_{c} = c_{1} x^{m_{1}} + c_{2} x^{m_{2}}$
Case 2 : ( $m_{1} = m_{2}$ )

use the method of reduction of order

$y_{2} (x) = y_{1} (x) \int \frac{e^{- \int P (x) d x}}{y_{1}^{2} (x)} d x$

$y_{1} (x) = x^{m_{1}}$

$y_{2} (x) = x^{m_{1}} \cdot \ln | x |$

直接背結論 $↑$
Case 3 : ( $D < 0$ )

$m_{1} = α + i β, m_{2} = α - i β$

直接背

$y_{c} = x^{α} [c_{1} \cos (β \ln x) + c_{2} \sin (β \ln x)]$

Solution to Higher Order¶

Case 1 : 皆唯一解

$x^{m_{p}}$
Case 2 : 在 $m_{p}$ 處有 $k$ 個重根

$x^{m_{0}}, x^{m_{0}} \ln x, x^{m_{0}} \cdot (\ln x)^{2}, \dots, x^{m_{0}} \cdot (\ln x)^{k - 1}$
Case 3 : 有一對複數根

$x^{α} \cos (β \ln x), x^{α} \sin (β \ln x)$
Case 4 : 有 $k$ 對複數解

$\begin{array}{r} x^{α} \cos (β \ln x), x^{α} \cos (β \ln x) \cdot \ln x, \dots, x^{α} \cos (β \ln x) \cdot (\ln x)^{k - 1} \\ x^{α} \sin (β \ln x), x^{α} \sin (β \ln x) \cdot \ln x, \dots, x^{α} \sin (β \ln x) \cdot (\ln x)^{k - 1} \end{array}$

Non-homogeneous Equations¶

Concept¶

we can solve any non-homogeneous equations by

y = y_{c} + y_{p}

in which $y_{c}$ is the complementary function and $y_{p}$ is the particular solution to equation.

Linear DE with Constant Coefficients¶

Condition¶

linear
constant coefficient
$g^{(k)} (k \in ℕ)$ contain finite number of terms.¹

Solution¶

Trial particular solutions

$g (x)$	Form of $y_{p}$
$const$	$A$
$5 x + 7$	$A x + B$
$3 x^{2} - 2$	$A x^{2} + B x + C$
$x^{3} - x + 1$	$A x^{3} + B x^{2} + C x + D$
$\sin 4 x$	$A \cos 4 x + B \sin 4 x$
$\cos 2 x$	$A \cos 2 x + B \sin 2 x$
$e^{5 x}$	$A e^{5 x}$
$(9 x - 2) e^{5 x}$	$(A x + B) e^{5 x}$
$x^{2} e^{5 x}$	$(A x^{2} + B x + C) e^{5 x}$
$x e^{3 x} \sin 4 x$	$(A x + B) e^{3 x} \cos 4 x + (C x + D) e^{3 x} \sin 4 x$

e.g.

g (x) = e^{2 x} + x e^{3 x}

\begin{aligned} y_{p_{1}} = A e^{2 x} & y_{p_{2}} = (B x + C) e^{3 x} \end{aligned}

y_{p} = y_{p_{1}} + y_{p_{2}} = A e^{2 x} + (B x + C) e^{3 x}

Glitch¶

Condition 1 : particular cannot belong to complementary function.

e.g.

$y^{″} - 5 y^{'} + 4 y = 8 e^{x}$

complementary function :

$y_{c} = c_{1} e^{x} + c_{2} e^{4 x}$

$\begin{matrix} ∵ A e^{x} \in y_{c} \end{matrix}$

therefore we guess the particular solution with an extra $x$ .

$y_{p} = A x e^{x}$
Condition2 :

$g (x), g^{'} (x), g^{″} (x), g^{‴} (x), \dots$

can only contain infinite number of terms.

Any Linear DE¶

a_{n} (x) y^{(n)} (x) + a_{n - 1} (x) y^{(n - 1)} + \dots + a_{1} (x) y^{'} (x) + a_{0} (x) y = g (x)

Variation of Parameters¶

Solution to 2nd Order

a_{2} (x) y^{″} (x) + a_{1} (x) y^{'} (x) + a_{0} (x) y (x) = g (x)

or in the form

y^{″} (x) + P (x) y^{'} + Q (x) y = f (x)

Suppose the complementary function is

c_{1} y_{1} (x) + c_{2} y_{2} (x)

then assume particular solution as

y_{p} = u_{1} (x) y_{1} (x) + u_{2} (x) y_{2} (x)

and then we can get

{\begin{array}{rc} y_{1} u_{1}^{'} + y_{2} u_{2}^{'} & = 0 \\ y_{1}^{'} u_{1}^{'} + y_{2}^{'} u_{2}^{'} & = f (x) \end{array}

\begin{aligned} u_{1}^{'} = \frac{W_{1}}{W} & u_{1} = \int u_{1} d x \\ u_{2}^{'} = \frac{W_{2}}{W} & u_{2} = \int u_{2} d x \end{aligned}

in which

\begin{aligned} W = Wronskian & = | \begin{array}{c} y_{1} & y_{2} \\ y_{1}^{'} & y_{2}^{'} \end{array} | \\ W_{1} & = | \begin{array}{c} 0 & y_{2} \\ f (x) & y_{2}^{'} \end{array} | \\ W_{2} & = | \begin{array}{c} y_{1} & 0 \\ y_{1}^{'} & f (x) \end{array} | \end{aligned}

ref : Wronskian

Higher Order¶

similar to the way above.

$y_{p}$ should be a linear combination of

$g (x), g^{'} (x), g^{″} (x), g^{‴} (x), \dots$

therefore, we won't get any redundant term. ↩