# on 16-Nov-2023 (Thu)

#### 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

#### Annotation 7597229739276

 The work it takes to move a charge in static E field Consider an electric field $$\vec{E}$$ and two points in space, $$a$$ and $$b$$, the work required to move a test charge $$Q$$ from $$a$$ to $$b$$ is $$W=\int^{b}_{a}\vec{F}\cdot d\vec{l}=-Q\int^{b}_{a}\vec{E}\cdot d\vec{l}$$ and becasue of the electrostatic field and the stokes's theorem, we have $$\oint_{P}\vec{E}\cdot d\vec{l}=\int_{S}(\nabla\times\vec{E})\cdot d\vec{a}=0$$ for any closed path.

#### Flashcard 7597233147148

Question

Consider an electric field $$\vec{E}$$ and two points in space, $$a$$ and $$b$$, the work required to move a test charge $$Q$$ from $$a$$ to $$b$$ is [...]

$$W=\int^{b}_{a}\vec{F}\cdot d\vec{l}=-Q\int^{b}_{a}\vec{E}\cdot d\vec{l}$$ and becasue of the electrostatic field and the stokes's theorem, we have $$\oint_{P}\vec{E}\cdot d\vec{l}=\int_{S}(\nabla\times\vec{E})\cdot d\vec{a}=0$$ for any closed path.

status measured difficulty not learned 37% [default] 0

The work it takes to move a charge in static E field
Consider an electric field $$\vec{E}$$ and two points in space, $$a$$ and $$b$$, the work required to move a test charge $$Q$$ from $$a$$ to $$b$$ is $$W=\int^{b}_{a}\vec{F}\cdot d\vec{l}=-Q\int^{b}_{a}\vec{E}\cdot d\vec{l}$$ and becasue of the electrostatic field and the stokes's theorem, we have $$\oint_{P}\vec{E}\cdot d\vec{l}=\int_{S}(\nabla\times\vec{E})\cdot d\vec{a}=0$$ for any closed path.

#### Annotation 7597234720012

 The Fundamental Theorem for Gradients For any vector field, $$\int^{b}_{a}(\nabla T)\cdot d\mathbf{l}=T(\mathbf{b})-T(\mathbf{a})$$

#### Flashcard 7597237341452

Question

For any vector field, [The Fundamental Theorem for Gradients]

$$\int^{b}_{a}(\nabla T)\cdot d\mathbf{l}=T(\mathbf{b})-T(\mathbf{a})$$

status measured difficulty not learned 37% [default] 0

For any vector field, $$\int^{b}_{a}(\nabla T)\cdot d\mathbf{l}=T(\mathbf{b})-T(\mathbf{a})$$

#### Annotation 7597240487180

 The Fundamental Theorem for Divergences For any vector field, region and boundary, $$\int_\mathcal{V}(\nabla\cdot\mathbf{v})d\tau=\oint_\mathcal{S}\mathbf{v}\cdot d\mathbf{a}$$

#### Flashcard 7597242322188

Question
For any vector field, region and boundary, [The Fundamental Theorem for Divergences]
$$\int_\mathcal{V}(\nabla\cdot\mathbf{v})d\tau=\oint_\mathcal{S}\mathbf{v}\cdot d\mathbf{a}$$

status measured difficulty not learned 37% [default] 0

The Fundamental Theorem for Divergences
For any vector field and boundary, $$\int_\mathcal{V}(\nabla\cdot\mathbf{v})d\tau=\oint_\mathcal{S}\mathbf{v}\cdot d\mathbf{a}$$

#### Annotation 7597244681484

 The Fundamental Theorem for Curls For any vector field, region and boundary, $$\int_\mathcal{S}(\nabla\times\mathbf{v})\cdot d\mathbf{a}=\oint_\mathcal{P}\mathbf{v}\cdot d\mathbf{l}$$

#### Flashcard 7597246516492

Question
For any vector field, region and boundary, [The Fundamental Theorem for Curls]
$$\int_\mathcal{S}(\nabla\times\mathbf{v})\cdot d\mathbf{a}=\oint_\mathcal{P}\mathbf{v}\cdot d\mathbf{l}$$
For any vector field, region and boundary, $$\int_\mathcal{S}(\nabla\times\mathbf{v})\cdot d\mathbf{a}=\oint_\mathcal{P}\mathbf{v}\cdot d\mathbf{l}$$