13 Operator Method
Operator Method is a more general method, so it is good.
Consider a 2nd order DE
\[ \begin{aligned} y'' + py' + qy &= R(x) \\ (D^2 + pD + q)y &= R(x) \end{aligned} \]
Definition¶
\[ y_p = \frac{1}{\phi(D)} R(x) \]
\[ \begin{aligned} \phi(D) y &= R(x) \\ \phi(D) &= D^2 + pD + q \\ &=(D-m_1)(D-m_2) \end{aligned} \]
Integrals¶
\[ \begin{aligned} \frac{1}{D} R(x) &= \int R(x) dx \\ \frac{1}{D^2} R(x) &= \iint R(x) dx \cdot dx \end{aligned} \]
\[ \begin{aligned} \frac{1}{D-m} R(x) &= \textcolor{orange}{e^{mx}} \int R(x) \cdot \textcolor{hotpink}{e^{-mx}} \cdot dx \\ \frac{1}{D+m} R(x) &= \textcolor{hotpink}{e^{-mx}} \int R(x) \cdot \textcolor{orange}{e^{mx}} \cdot dx \end{aligned} \]
Short Rules for standard functions¶
| R(x) | \(y_p\) | Exception |
|---|---|---|
| \(e^{\textcolor{orange}{a}x}\) | \(e^{\textcolor{orange}{a}x}\dfrac{1}{\phi(\textcolor{orange}{a})}\) | |
| \(x e^{\textcolor{orange}{a}x} \frac{1}{\phi'(\textcolor{orange}{a})}\) | \(\phi(a) = 0\) | |
| \(x^2 e^{\textcolor{orange}{a}x} \frac{1}{\phi''(\textcolor{orange}{a})}\) | \(\phi'(a) = 0\) | |
| \(\sin(\textcolor{orange}{a}x+b), \cos(\textcolor{orange}{a}x+b)\) | \(R(x) \frac{1}{f(-\textcolor{orange}{a}^2)}\) | |
| \(x R(x) \frac{1}{f'(-\textcolor{orange}{a}^2)}\) | \(f(-a^2) = 0\) | |
| \(x^2 R(x) \frac{1}{f''(-\textcolor{orange}{a}^2)}\) | \(f'(-a^2) = 0\) | |
| \(x^m\) | \(\underbrace{\Big(\phi(D) \Big)^{-1} }_\text{Binomial Series} x^m\) | |
| \(e^{\textcolor{orange}{k}x} h(x)\) (Exponent Shifting Rule) | \(e^{\textcolor{orange}{k}x} \left\{ \frac{1}{\phi(D+\textcolor{orange}{k})} h(x) \right\}\) Solve using any of the above methods |
Derivatives¶
\[ \begin{aligned} \phi'(a) &= \left\{ \frac{d \phi(D)}{dD} \right\}_{D \to a} \\ \phi''(a) &= \left\{ \frac{d^2 \phi(D)}{d D^2} \right\}_{D \to a} \\ f(-a^2) &= \left\{ \frac{d f(D^2)}{dD} \right\}_{D^2 \to -a^2} \\ f'(-a^2) &= \left\{ \frac{d^2 f(D^2)}{d D^2} \right\}_{D^2 \to -a^2} \\ \end{aligned} \]
Binomial Expansions¶
\[ \begin{aligned} (1+x)^{-1} &= 1 - x + x^2 - x^3 + \dots \\ (1-x)^{-1} &= 1 + x + x^2 + x^3 + \dots \end{aligned} \]
\[ \begin{aligned} (1+x)^{-2} &= 1 - 2x + 3x^2 - 4x^3 + \dots \\ (1-x)^{-2} &= 1 + 2x + 3x^2 + 4x^3 + \dots \end{aligned} \]
\[ \begin{aligned} (1+x)^{-n} &= 1 - nx + \frac{n(n+1) x^2}{2!} - \frac{n(n+1)(n+2) x^3}{3!} + \cdots\\ (1-x)^{-n} &= 1 + nx + \frac{n(n+1) x^2}{2!} + \frac{n(n+1)(n+2) x^3}{3!} + \cdots \end{aligned} \]
Cube Formula¶
\[ (a+b)^3 = a^3+b^3+3ab(a+b) \\ (a-b)^3 = a^3-b^3-3ab(a-b) \]
Long Method¶
idk¶
\[ \begin{aligned} y_p &= \frac{1}{\phi(D)} R(x) \\ &= \underbrace{ \left( \frac{1}{D-m_1} \right) \underbrace{ \frac{1}{D-m_2} R(x) }_{R_1(x)} }_{R_2(x)}\\ &= \frac{1}{D-m} R(x) \end{aligned} \]
IF¶
\[ \begin{aligned} Dy - my &= R(x) \\ \frac{dy}{dx} - my &= R(x) \\ IF &= e^{\int P(x) dx} \\ &= e^{\int -m dx} \\ &= e^{-mx} \end{aligned} \]
Solution¶
\[ \begin{aligned} y \times IF &= \int R(x) \cdot IF \cdot dx \\ y e^{-mx} &= \int R(x) \cdot e^{-mx} \cdot dx \\ y &= e^{mx} \int R(x) \cdot e^{-mx} \cdot dx \end{aligned} \]