# on 15-Nov-2023 (Wed)

#### 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 7597097618700

Question
For the non-constant coefficients linear homogeneous ODE, there are only two solution methods: [Answer the method name and their applied condition]

- If the ODE is like $$x^{2}y''+bxy'+cy=0$$, then it can be fully solved: "Euler Equation" Method
- Otherwise, need "Red of Order"

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The solution method for non-constant-coefficients 2nd order linear homogeneous ODE
For the non-constant coefficients linear homogeneous ODE, there are only two solution methods: - If the ODE is like $$x^{2}y''+bxy'+cy=0$$, then it can be fully solved: "Euler Equation" Method - Otherwise, need "Red of Order"

#### Annotation 7597203787020

 General form for 2nd Linear Inhomogeneous ODE The general form of 2nd order linear inhomogeneous ODE is $$\frac{d^{2}y}{dx^{2}}+p(x)\frac{dy}{dx}+q(x)y(x)=r(x)$$, where $$p(x),q(x),r(x)$$ are the function of $$x$$.

#### Flashcard 7597206932748

Question
The general form of 2nd order linear inhomogeneous ODE is [...], where $$p(x),q(x),r(x)$$ are the function of $$x$$.
$$\frac{d^{2}y}{dx^{2}}+p(x)\frac{dy}{dx}+q(x)y(x)=r(x)$$

status measured difficulty not learned 37% [default] 0

General form for 2nd Linear Inhomogeneous ODE
The general form of 2nd order linear inhomogeneous ODE is $$\frac{d^{2}y}{dx^{2}}+p(x)\frac{dy}{dx}+q(x)y(x)=r(x)$$, where $$p(x),q(x),r(x)$$ are the function of $$x$$.

#### Annotation 7597208505612

 Main idea to solve 2nd order linear inhomogeneous ODE For 2nd Order Linear Inhomogeneous ODE, we need to find its complementary functions and its particular integrals. Getting its complementary functions by solving the homogeneous ODE constructed by replacing $$r(x)$$ with 0 to get its complementary functions. Getting its particular integrals using the "Variation of Parameters" method.

#### Flashcard 7597212962060

Question
For solving the 2nd Order Linear Inhomogeneous ODE, we need to find [Find what and how to get them]
its complementary functions and its particular integrals. Getting its complementary functions by solving the homogeneous ODE constructed by replacing $$r(x)$$ with 0 to get its complementary functions. Getting its particular integrals using the "Variation of Parameters" method.

status measured difficulty not learned 37% [default] 0

Main idea to solve 2nd order linear inhomogeneous ODE
For 2nd Order Linear Inhomogeneous ODE, we need to find its complementary functions and its particular integrals. Getting its complementary functions by solving the homogeneous ODE constructed by replacing $$r(x)$$ with 0 to get its complementary functions. Getting its particular integrals using the "Variation of Parameters" method.

#### Annotation 7597217418508

 Variation of Parameters for getting the particular integral of a 2nd order linear homogeneous ODE For a 2nd order linear inhomogeneous ODE $$\frac{d^{2}y}{dx^{2}}+p(x)\frac{dy}{dx}+q(x)y(x)=r(x)$$ The particular integral of it is given by $$y_{p}(x)=u_{2}(x)\int^{x}\frac{u_{1}(\zeta)r(\zeta)}{W[u_{1}(\zeta),u_{2}(\zeta)]}d\zeta-u_{1}(x)\int^{x}\frac{u_{2}(\zeta)r(\zeta)}{W[u_{1}(\zeta),u_{2}(\zeta)]}d\zeta=\int^x \frac{\left|\begin{array}{cc}u_1(\zeta) & u_2(\zeta) \\u_1(x) & u_2(x)\end{array}\right|}{W\left[u_1(\zeta), u_2(\zeta)\right]} r(\zeta) d \zeta$$ where: - $$\displaystyle W[u_{1}(x),u_{2}(x)]=\begin{vmatrix}u_{1}(x) & u_{2}(x) \\ u_{1}'(x) & u_{2}'(x)\end{vmatrix}=u_{1}(x)u_{2}'(x)-u_{2}(x)u_{1}'(x)$$. - $$u_{1}(x), u_{2}(x)$$ is the complementary functions of the inhomogeneous ODE. - $$r(x)$$ is RHS of the inhomogeneous ODE.

#### Flashcard 7597220039948

Question

For a 2nd order linear inhomogeneous ODE

$$\frac{d^{2}y}{dx^{2}}+p(x)\frac{dy}{dx}+q(x)y(x)=r(x)$$
The particular integral of it is given by[Using "Varient of Parameters" and show all forms of the formula]

$$y_{p}(x)=u_{2}(x)\int^{x}\frac{u_{1}(\zeta)r(\zeta)}{W[u_{1}(\zeta),u_{2}(\zeta)]}d\zeta-u_{1}(x)\int^{x}\frac{u_{2}(\zeta)r(\zeta)}{W[u_{1}(\zeta),u_{2}(\zeta)]}d\zeta=\int^x \frac{\left|\begin{array}{cc}u_1(\zeta) & u_2(\zeta) \\u_1(x) & u_2(x)\end{array}\right|}{W\left[u_1(\zeta), u_2(\zeta)\right]} r(\zeta) d \zeta$$

where:
- $$\displaystyle W[u_{1}(x),u_{2}(x)]=\begin{vmatrix}u_{1}(x) & u_{2}(x) \\ u_{1}'(x) & u_{2}'(x)\end{vmatrix}=u_{1}(x)u_{2}'(x)-u_{2}(x)u_{1}'(x)$$.
- $$u_{1}(x), u_{2}(x)$$ is the complementary functions of the inhomogeneous ODE.
- $$r(x)$$ is RHS of the inhomogeneous ODE.

status measured difficulty not learned 37% [default] 0

Variation of Parameters for getting the particular integral of a 2nd order linear homogeneous ODE
For a 2nd order linear inhomogeneous ODE $$\frac{d^{2}y}{dx^{2}}+p(x)\frac{dy}{dx}+q(x)y(x)=r(x)$$ The particular integral of it is given by $$y_{p}(x)=u_{2}(x)\int^{x}\frac{u_{1}(\zeta)r(\zeta)}{W[u_{1}(\zeta),u_{2}(\zeta)]}d\zeta-u_{1}(x)\int^{x}\frac{u_{2}(\zeta)r(\zeta)}{W[u_{1}(\zeta),u_{2}(\zeta)]}d\zeta=\int^x \frac{\left|\begin{array}{cc}u_1(\zeta) & u_2(\zeta) \\u_1(x) & u_2(x)\end{array}\right|}{W\left[u_1(\zeta), u_2(\zeta)\right]} r(\zeta) d \zeta$$ where: - $$\displaystyle W[u_{1}(x),u_{2}(x)]=\begin{vmatrix}u_{1}(x) & u_{2}(x) \\ u_{1}'(x) & u_{2}'(x)\end{vmatrix}=u_{1}(x)u_{2}'(x)-u_{2}(x)u_{1}'(x)$$. - $$u_{1}(x), u_{2}(x)$$ is the complementary functions of the inhomogeneous ODE. - $$r(x)$$ is RHS of the inhomogeneous ODE.

#### Annotation 7597222923532

 Definition of Electrostatic Field In electrostatics, we ask $$\vec{E}$$ to be independent of time, but it can change with position. $$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_{0}}$$ We also ask $$\vec{B}$$ to be independent of time. $$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}=0$$ - This equation is used to decide whether an electric field can be an electrostatic field

#### Flashcard 7597225282828

Question
In electrostatics, we ask [$$\vec{E}$$?]
We also ask [$$\vec{B}$$?]
- This equation is used to decide whether an electric field can be an electrostatic field

$$\vec{E}$$ to be independent of time, but it can change with position.

$$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_{0}}$$

$$\vec{B}$$ to be independent of time.

$$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}=0$$

status measured difficulty not learned 37% [default] 0

Definition of Electrostatic Field
In electrostatics, we ask $$\vec{E}$$ to be independent of time, but it can change with position. $$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_{0}}$$ We also ask $$\vec{B}$$ to be independent of time. $$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}=0$$ - This equation is used to decide whether an electric field can be an el