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Depending on the solution of the auxiliary equation, the solution of the ODE is different: \(m=\frac{-b\pm\Delta}{2a}, \Delta\equiv b^{2}-4ac\) 1. \(\Delta>0\Rightarrow\) 2 real roots: \(m_{1},m_{2}\), both \(y_{1}=e^{m_{1}x},y_{2}=e^{m_{2}x}\) works. So the general solution is \(y_{general}=Ae^{m_{1}x}+Be^{m_{2}x}\) - Alternative form: Because of the Euler's Equation in \(\mathbb{C}\), \(y(x)=e^{px}[\tilde{A}\cosh(qx)+\tilde{B}\sinh(qx)]\), where \(p,q\) are \(\displaystyle m=\underbrace{- \frac{b}{2a}}_p\pm \underbrace{\frac{\sqrt{\Delta}}{2a}}_{q}\). - Why we need this form? When the initial condition includes \(y(0)\), the alt form would be much easier 2. \(\Delta<0\Rightarrow\) 2 complex roots: \(m_{1},m_{2}\in\mathbb{C}\), so the general solution is \(y(x)=Ce^{m_{1}x}+De^{m_{2}x}\) - Alternative form: Because of the Euler's Equation, \(y(x)=e^{m_{r}x}[\tilde{C}\cosh(m_{i}x)+\tilde{D}\sinh(m_{i}x)]\), where \(m_{1,2}=m_{r}\pm im_{i}\). 3. \(\Delta=0\Rightarrow\) one \(m\) only! so the general solution is \(y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}\)

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The Taylor series of a infinitely differentiable function \(f(x)\in\mathbb{C}\) at a single point \(a\in\mathbb{C}\) is defined as: \({\displaystyle f(a)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}}\) where \(f^{(n)}(a)\) denotes the \(n\)th derivative of \(f\) evaluated at the point \(a\). - The derivative of order zero of \(f\) is defined to be \(f\) itself and \((x − a)^{0}\) and \(0!\) are both defined to be 1.

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The Fourier series of a 2L-periodic function \(f(x)\) is defined as: \(\displaystyle f(x)=\frac{a_{0}}{2}+\sum\limits^{\infty}_{n=1}a_{n}\cos\left(\frac{n\pi x}{L}\right)+b_{n}\sin\left(\frac{n\pi x}{L}\right)\) where - \(\displaystyle a_{0}=\frac{1}{L}\int^{L}_{-L}f(x)dx\equiv \left\langle1,f\right\rangle\). - \(\displaystyle a_{n}=\frac{1}{L}\int^{L}_{-L}f(x)\cos\left(\frac{n\pi x}{L}\right)dx\equiv \frac{\left\langle\cos_{n},f\right\rangle}{\left\langle\cos_{n},\cos_{n}\right\rangle}\). - \(\displaystyle b_{n}=\frac{1}{L}\int^{L}_{-L}f(x)\sin\left(\frac{n\pi x}{L}\right)dx\equiv \frac{\left\langle\sin_{n},f\right\rangle}{\left\langle\sin_{n},\sin_{n}\right\rangle}\).

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For specific weight funciton \(w(x)=1\) and domain \(x\in(-L,L)\), then - \(\displaystyle\left\langle \cos\left(\frac{n\pi x}{L}\right), \cos\left(\frac{m\pi x}{L}\right)\right\rangle=\int^{L}_{-L}\cos\left(\frac{n\pi x}{L}\right)\ \cos\left(\frac{m\pi x}{L}\right)\cdot1dx=L\cdot\delta_{nm}\) - \(\displaystyle\left\langle \sin\left(\frac{n\pi x}{L}\right), \sin\left(\frac{m\pi x}{L}\right)\right\rangle=\int^{L}_{-L}\sin\left(\frac{n\pi x}{L}\right)\ \sin\left(\frac{m\pi x}{L}\right)\cdot1dx=L\cdot\delta_{nm}\) - \(\displaystyle\left\langle \cos\left(\frac{n\pi x}{L}\right), \sin\left(\frac{m\pi x}{L}\right)\right\rangle=\int^{L}_{-L}\cos\left(\frac{n\pi x}{L}\right)\ \sin\left(\frac{m\pi x}{L}\right)\cdot1dx=0\)