Higher Order DE

homogeneous Equations¶

Definition¶

or be written as

• \(g(x) = 0 \qquad \rightarrow \quad\) homogeneous
• \(g(x) \neq 0 \qquad \rightarrow \quad\) 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

  \[y=c_1y_1 + c_2y_2 + \dots + c_ny_n\]

• 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)

2nd with an known solution¶

2^nd order linear homogeneous DE with an known solution

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

Conditions¶

1. second order
2. linear
3. homogeneous
4. one of the nontrivial solution \(y_1(x)\) has been known

#### Standard Form

Solution¶

(直接背！！！)

see proof. (TODO)

Linear DE with Constant Coefficients¶

Homogeneous linear DE with constant coefficients

Condition¶

1. homogeneous
2. linear
3. constant coefficients

kernel concept¶

Suppose the solutions has the form of \(e^{mx}\)

Auxiliary Function¶

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

solve \(m\).

Solution to 2nd Order¶

• Case 1 : \(m_1 \neq m_2, \quad m_1, m_2 \in ℝ\), (D > 0)

  \[m_1, m_2 = \frac{-a_1 \pm \sqrt{a_1^2-4a_2a_0}}{2a_2}\]

  thus

  \[y = c_1e^{m_1x} + c_2e^{m_2x}\]

  if

  \[ m_1, m_2 = \alpha \ \pm \ \beta \]

  we can also write

  \[ y=e^{\alpha x}(c_1\cosh{\beta x} + c_2\sinh{\beta x}) \]

• Case 2 \(m_1 = m_2\), (D = 0)

  \[y_1 = e^{m_1x}\]

  \(y_2\) can be find by the method mentioned above.

  then we found that

  \[y_2 = x\ e^{m_1x}\]

  
  

• Case 3 \(m_1 \neq m_2\), \(m_1\) and \(m_2\) are conjugate(共軛) and complex, (D < 0)

  \[ \begin{gather} m_1 = \alpha + i\beta \qquad m_2 = \alpha - i\beta \\ \\ \alpha = -\frac{a_1}{2a_2}, \qquad \beta = \frac{\sqrt{4a_2a_0 - a_1^2}}{2a_2} \end{gather} \]

  thus

  \[ \begin{gather} y = C_1e^{\alpha x + i\beta x} + C_2 e^{\alpha x - i\beta x} \end{gather} \]

  another form : proof

  \[y = e^{\alpha x}(c_1 \cos{\beta x} + c_2 \sin{\beta x})\]

Solution to Higher Order¶

n^th order ODE

\(p, q \in [1, n], \quad \text{and} \quad p \neq q\)

• 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 : \(\quad e^{m_px},\quad x\cdot e^{m_px},\quad x^2\cdot e^{m_px}, \dots ,\quad x^{k-1}\cdot e^{m_px}\)

• Case 3 : 有k對複數解

  solutions :

  \[ \begin{align} e^{\alpha x} \cos(\beta x),\quad xe^{\alpha x} \cos(\beta x),\quad x^2e^{\alpha x} \cos(\beta x),\dots , x^{k-1}e^{\alpha x}\cos (\beta x) \\\\ e^{\alpha x} \sin(\beta x),\quad xe^{\alpha x} \sin(\beta x),\quad x^2e^{\alpha x} \sin(\beta x),\dots , x^{k-1}e^{\alpha x}\sin (\beta x) \end{align} \]

  ---

Cauchy-Euler Equation¶

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

Kernel Concept¶

guess the solution has the form \(y(x) = x^m\)

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

Solution to 2nd Order¶

• auxiliary function :

  \[ \begin{align} a_2m(m-1) &+ a_1m &+ a_0 &= 0\\\\ a_2m^2 &+ (a_1 - a_2)m &+ a_0 &= 0 \end{align} \]

than solve roots with

• Case 1 : ( \(m_1 \neq m_2, \quad m_1, m_2 \in ℝ\) )

  \[y_c = c_1x^{m_1} + c_2 x^{m_2}\]

• Case 2 : ( \(m_1 = m_2\) )

  use the method of reduction of order

  \[y_2(x) = y_1(x) \int{\frac{e^{-\int P(x)dx}}{y_1^2(x)}dx}\]

  \[y_1(x) = x^{m_1}\]

  \[y_2(x) = x^{m_1} \cdot \ln{|x|}\]

  直接背結論 \(\uparrow\)

• Case 3 : ( \(D < 0\) )

  \[m_1 = \alpha + i\beta, \qquad m_2 = \alpha - i\beta\]

  直接背

  \[y_c = x^{\alpha}\big[c_1\cos{(\beta \ln x)} + c_2 \sin{(\beta \ln x)}\big]\]

  ---

Solution to Higher Order¶

• Case 1 : 皆唯一解

  \[x^{m_p}\]

• Case 2 : 在\(m_p\)處有 \(k\) 個重根

  \[x^{m_0}, \quad x^{m_0}\ln x,\quad x^{m_0}\cdot (\ln x)^2, \dots ,\quad x^{m_0}\cdot (\ln x)^{k-1} \]

• Case 3 : 有一對複數根

  \[x^{\alpha}\cos{(\beta \ln x)}, \quad x^{\alpha}\sin{(\beta \ln x)} \]

• Case 4 : 有 \(k\) 對複數解

  \[ \begin{align} x^{\alpha}\cos{(\beta \ln x)}, \quad x^{\alpha}\cos{(\beta \ln x)} \cdot \ln x \ , \dots,\ x^{\alpha}\cos{(\beta \ln x)}\cdot (\ln x)^{k-1} \\\\ x^{\alpha}\sin{(\beta \ln x)}, \quad x^{\alpha}\sin{(\beta \ln x)}\cdot \ln x\ , \dots,\ x^{\alpha}\sin{(\beta \ln x)}\cdot (\ln x)^{k-1} \end{align} \]

Non-homogeneous Equations¶

Concept¶

we can solve any non-homogeneous equations by

in which \(y_c\) is the complementary function and \(y_p\) is the particular solution to equation.

Linear DE with Constant Coefficients¶

Condition¶

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

Solution¶

Trial particular solutions

e.g.

Glitch¶

• Condition 1 : particular cannot belong to complementary function.

  e.g.

  \[y'' -5y' +4y= 8e^x\]

  complementary function :

  \[ y_c = c_1e^x + c_2e^{4x}\]
  \[ \begin{gather} \because Ae^{x} \in y_c \end{gather} \]

  therefore we guess the particular solution with an extra \(x\).

  \[y_p = Axe^x\]

• Condition2 :

  \[g(x), \ g'(x), \ g''(x), \ g'''(x), \dots\]

  can only contain infinite number of terms.

Any Linear DE¶

Variation of Parameters¶

Solution to 2nd Order

or in the form

Suppose the complementary function is

then assume particular solution as

and then we can get

in which

ref : Wronskian

Higher Order¶

similar to the way above.