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The order of a differential equation is the highest derivative within the equation.

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Depending on whether the ODE has non-linear terms, the classification of the ODE are Linear ODE and Non-linear ODE.

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Depending on whether the ODE has non-linear terms, the classification of the ODE are Linear ODE and Non-linear ODE.

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The general Form for \(2^{nd}\) Order Linear ODE is \(a(x)\frac{d^{2}y}{dx^{2}}+b(x)\frac{dy}{dx}+c(x)y=d(x)\) where \(a(x),b(x),c(x),d(x)\) are different function of \(x\).

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The homogeneous ODE is the ODE which has no terms with a pure function of \(x\).

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The inhomogeneous ODE is the ODE which has at least one term with a pure function of \(x\).

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Depending on whether it has a pure function of non-variable function \(x\), the classification of the ODE is the homogeneous ODE(Having no non-variable function) and inhomogeneous ODE(Having at least one non-variable function).

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If an ODE is linear and homogeneous, then the "Principle of Superposition" appears: If \(y_{1}\) and \(y_{2}\) both solve an ODE, then, \(y_{3}=\alpha y_{1}+\beta y_{2}\) is also a solution.

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If an ODE is linear and homogeneous, then the "Principle of Superposition" appears: If \(y_{1}\) and \(y_{2}\) both solve an ODE, then, \(y_{3}=\alpha y_{1}+\beta y_{2}\) is also a solution.

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The general form of \(2^{nd}\) order linear homogeneous ODE is \(a_{2}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y(x)=0\) where \(a_{n}(x),\ n=0,1,2\) are the function of \(x\).

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欧拉公式提出，对任意实数\(x\)，都存在\(e^{ix} = \cos x + i\sin x\)，其中\(e\)是自然对数的底数，\(i\)是虚数单位，而\(\cos\)和\(\sin\)则是余弦、正弦对应的三角函数，参数\(x\)则以弧度为单位

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由于该公式在\(x\)为复数时仍然成立，所以也有人将这一更通用的版本称为欧拉公式：\(e^{iθ} = \cosθ + i\sin θ \iff e^θ = \cosh θ + \sinh θ\)

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> The general form of 2nd-order linear homogeneous ODE is \(a_{2}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y(x)=0\) If \(a_{2}(x),a_{1}(x),a_{0}(x)\) are constant functions: 1. Try ansatz: \(y=e^{mx}\). 2. To find \(m\), plug \(y = e^{mx}\) into ODE. Rearrange the equation and get the auxiliary equation: \(\underbrace{(am^{2}+bm+c)}_{\text{Auxiliary Equation}}e^{mx}=0\) 3. Using the solution of the auxiliary equation to solve the ODE. <span>

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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 \(y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}\)