Edited, memorised or added to reading queue

on 16-Nov-2023 (Thu)

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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]

Answer

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

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.


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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
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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.
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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 [...]

Answer
\(W=\int^{b}_{a}\vec{F}\cdot d\vec{l}=-Q\int^{b}_{a}\vec{E}\cdot d\vec{l}\) 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.

statusnot learnedmeasured difficulty37% [default]last interval [days]               
repetition number in this series0memorised on               scheduled repetition               
scheduled repetition interval               last repetition or drill

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, \(\int^{b}_{a}(\nabla T)\cdot d\mathbf{l}=T(\mathbf{b})-T(\mathbf{a})\)
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Flashcard 7597237341452

Question

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

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

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The Fundamental Theorem for Gradients
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, \(\int_\mathcal{V}(\nabla\cdot\mathbf{v})d\tau=\oint_\mathcal{S}\mathbf{v}\cdot d\mathbf{a}\)
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Flashcard 7597242322188

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

statusnot learnedmeasured difficulty37% [default]last interval [days]               
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scheduled repetition interval               last repetition or drill

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, \(\int_\mathcal{S}(\nabla\times\mathbf{v})\cdot d\mathbf{a}=\oint_\mathcal{P}\mathbf{v}\cdot d\mathbf{l}\)
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Flashcard 7597246516492

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

statusnot learnedmeasured difficulty37% [default]last interval [days]               
repetition number in this series0memorised on               scheduled repetition               
scheduled repetition interval               last repetition or drill

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