# on 08-Nov-2023 (Wed)

#### Annotation 7596175133964

 The principle of fault modeling is to reduce the number of effects to be tested by considering how defects manifest themselves.

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#### Annotation 7596977556748

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

#### Flashcard 7596978867468

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

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

#### Annotation 7596980964620

 Definition of Linear Ordinary Differential Equation The Linear Ordinary Differential Equation is the ordinary differential equation which has no non-linear terms like $$y^{2},y^{3},\cdots,y^{\prime 2},y^{\prime 3},\cdots,yy^{\prime},\cdots,y^{\prime}y^{\prime\prime},\cdots$$.

#### Annotation 7596984372492

 Definition of Non-linear Ordinary Differential Equation The Non-Linear Ordinary Differential Equation is the ordinary differential equation which has at least one non-linear terms like $$y^{2},y^{3},\cdots,y^{\prime 2},y^{\prime 3},\cdots,yy^{\prime},\cdots,y^{\prime}y^{\prime\prime},\cdots$$

#### Annotation 7596992498956

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

#### Flashcard 7596995906828

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

#### Flashcard 7596997479692

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

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

#### Flashcard 7597005606156

Tags
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Question

Which of the following ordinary differential equations are linear differential equations?

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#### Annotation 7597013994764

 General Form for $2^{nd}$ Order Linear ODE 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 $$x$$.

#### Flashcard 7597018189068

Question

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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General Form for $2^{nd}$ Order Linear ODE
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 $$x$$.

#### Annotation 7597020286220

 Definition of the homogeneous ODE The homogeneous ODE is the ODE which has no terms with a pure function of $$x$$.

#### Flashcard 7597021596940

Question

The homogeneous ODE is the ODE which [...].

has no terms with a pure function of $$x$$

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Definition of the homogeneous ODE
The homogeneous ODE is the ODE which has no terms with a pure function of $$x$$.

#### Annotation 7597023694092

 Definition of the inhomogeneous ODE The inhomogeneous ODE is the ODE which has at least one term with a pure function of $$x$$.

#### Flashcard 7597025529100

Question

The inhomogeneous ODE is the ODE which [...]

has at least one term with a pure function of $$x$$.

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Definition of the inhomogeneous ODE
The inhomogeneous ODE is the ODE which has at least one term with a pure function of $$x$$.

#### Annotation 7597027101964

 Classification of ODE by Homogeneity 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).

#### Flashcard 7597028936972

Question

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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Classification of ODE by Homogeneity
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).

#### Annotation 7597036539148

 Principle of Superposition for ODE's solutions 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.

#### Flashcard 7597038374156

Question
If an ODE is linear and homogeneous, then the "Principle of Superposition" appears: [The property of the solutions]
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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Principle of Superposition for ODE's solutions
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.

#### Flashcard 7597039947020

Question

If an ODE is [...], 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.

linear and homogeneous

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Principle of Superposition for ODE's solutions
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.

#### Annotation 7597042044172

 General Form for 2nd Oder Linear Homogeneous ODE 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 $$x$$.

#### Flashcard 7597043354892

Question
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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General Form for 2nd Oder Linear Homogeneous ODE
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 $$x$$.

#### Annotation 7597045714188

 Definition of Euler's formula 欧拉公式是复分析领域的公式，它将三角函数与复指数函数关联起来，因其提出者莱昂哈德·欧拉而得名

#### Flashcard 7597047549196

Question

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Definition of Euler's formula

#### Flashcard 7597049122060

Question

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Definition of Euler's formula

#### Annotation 7597050694924

 Mathematical Definition of Euler's Equation 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

#### Flashcard 7597053578508

Question

$$e^{ix} = \cos x + i\sin x$$

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Mathematical Definition of Euler's Equation

#### Annotation 7597055675660

 Euler's Equation in Complex number set 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：$$e^θ = \cosh θ + \sinh θ$$

#### Flashcard 7597061180684

Question

$$e^θ = \cosh θ + \sinh θ$$

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Euler's Equation in Complex number set

#### Annotation 7597063802124

 The solution method for Constant-coefficients 2nd order linear homogeneous ODE 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.

#### Flashcard 7597065637132

Question
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: [Method Steps]

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.

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The solution method for 2nd order linear homogeneous ODE
> 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>

#### Annotation 7597067996428

 The relations between the solution of an ODE and the solution of the ODE's auxiliary equation Depending on the solution of the auxiliary equation, the solution of the ODE is different: $$\displaystyle m=\frac{-b\pm\Delta}{2a}, \Delta\equiv b^{2}-4ac$$ 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, $$y(x)=e^{m_{r}x}[\tilde{C}\cos(m_{i}x)+\tilde{D}\sin(m_{i}x)]$$, where $$m_{1,2}=m_{r}\pm im_{i}$$. 3. $$\Delta=0\Rightarrow$$ one $$m$$ only! so the general solution is $$y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}$$

#### Flashcard 7597070617868

Question

Depending on the solution of the auxiliary equation, the solution of the corresponding 2nd Linear & homogeneous ODE is different: $$\displaystyle m=\frac{-b\pm\sqrt{\Delta}}{2a}, \Delta\equiv b^{2}-4ac$$

[Answer the normal solutions and their alternative form]

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, $$y(x)=e^{m_{r}x}[\tilde{C}\cos(m_{i}x)+\tilde{D}\sin(m_{i}x)]$$, where $$m_{1,2}=m_{r}\pm im_{i}$$.
3. $$\Delta=0\Rightarrow$$ one $$m$$ only! so the general solution is $$y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}$$
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 $$y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}$$