on 15-Nov-2023 (Wed)

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.

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 \(x\).

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.

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.

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