Added and updated ODE section.

config/en/mkdocs.yaml

@@ -58,6 +58,7 @@ nav:
       - 'First order differential equations': mathematics/ordinary-differential-equations/first-order-ode.md
       - 'Second order differential equations': mathematics/ordinary-differential-equations/second-order-ode.md
       - 'Systems of linear differential equations': mathematics/ordinary-differential-equations/systems-of-linear-ode.md
+      - 'The Laplace transform': mathematics/ordinary-differential-equations/laplace-transform.md
 
   - 'Physics':
     - 'Start': physics/start.md

docs/en/mathematics/ordinary-differential-equations/laplace-transform.md

@@ -0,0 +1 @@
+# The Laplace transform

docs/en/mathematics/ordinary-differential-equations/second-order-ode.md

@@ -40,7 +40,7 @@ $$
 L[y(t)] = \lambda^2 e^{\lambda t} + p \lambda e^{\lambda t} + q e^{\lambda t} = e^{\lambda t} (\lambda^2 + p \lambda + q) = 0,
 $$
 
-obtaining the characteristic equation $\Chi(\lambda) = \lambda^2 + p \lambda + q = 0$. If two roots $\lambda_1,\lambda_2 \in \mathbb{C}$ are found the solution space is 
+obtaining the characteristic equation $\chi(\lambda) = \lambda^2 + p \lambda + q = 0$. If two roots $\lambda_1,\lambda_2 \in \mathbb{C}$ are found the solution space is 
 
 $$
 y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}, \quad c_1,c_2 \in \mathbb{C},
@@ -56,7 +56,7 @@ $$
 
 #### Example
 
-Let the homogeneous linear second-order ode be given by $\ddot y + 4 \dot y + 8y = 0$. Then the characteristic equation is given by $\Chi(\lambda) = \lambda^2 + 4\lambda + 8 = 0$ with solutions $\lambda_1 = -2 + 2i$ and $\lambda_2 = -2 - 2i$. Then the general solution is given by
+Let the homogeneous linear second-order ode be given by $\ddot y + 4 \dot y + 8y = 0$. Then the characteristic equation is given by $\chi(\lambda) = \lambda^2 + 4\lambda + 8 = 0$ with solutions $\lambda_1 = -2 + 2i$ and $\lambda_2 = -2 - 2i$. Then the general solution is given by
 
 $$
 y(t) = c_1 e^{(-2 + 2i)1 t} + c_2 e^{(-2 - 2i) t}, \quad c_1,c_2 \in \mathbb{C},
@@ -118,7 +118,7 @@ named the Wronskian and we can solve for $c_1(t)$ and $c_2(t)$ by integration.
 
 #### Ansatz method
 
-Let $f(t) = p(t)e^{\lambda t}$, rule of thumb: $y_p$ is of related type to inhomogeneity $f$. Then for $A_n, B_n and P_n$ polynomials of degree $\leq n$ and $\alpha \in \R$
+Let $f(t) = p(t)e^{\lambda t}$, rule of thumb: $y_p$ is of related type to inhomogeneity $f$. Then for $A_n, B_n and P_n$ polynomials of degree $\leq n$ and $\alpha \in \mathbb{R}$
 
 | Inhomogeneity | Particular solution |
 | ------ | --------------- |
@@ -129,5 +129,5 @@ Let $f(t) = p(t)e^{\lambda t}$, rule of thumb: $y_p$ is of related type to inhom
 | $L[y] = P_n e^{\alpha t} \cos \omega t$ | $t^m e^{\alpha t} \big(A_n \cos \omega t + B_n \sin \omega t \big)$ |
 | $L[y] = P_n e^{\alpha t} \sin \omega t$ | $t^m e^{\alpha t} \big(A_n \cos \omega t + B_n \sin \omega t \big)$ |
 
-Choose $m \in \N \cup \{0\}$ as small as possible such that no term in the ansatz solves the homogeneous equation $L[y] = 0$.
+Choose $m \in \mathbb{N} \cup \{0\}$ as small as possible such that no term in the ansatz solves the homogeneous equation $L[y] = 0$.
 

docs/en/mathematics/ordinary-differential-equations/systems-of-linear-ode.md

@@ -1,2 +1,75 @@
 # Systems of linear ordinary differential equations
 
+## Homogeneous systems of linear ODEs with constant coefficients
+
+Let $\mathbb{K} = \mathbb{R} \lor \mathbb{C}$, $n \in \mathbb{N}$ and $A \in \mathbb{R}^{n \times n}$. Seek differentiable functions $y:\mathbb{R} \to \mathbb{K}^n$ such that 
+
+$$
+\mathbf{\dot y(t)} = A \mathbf{y}(t), \qquad t \in \mathbb{R}
+$$
+
+The solutions from a linear space, therefore the general solutions can be written as,
+
+$$
+\mathbf{\dot y(t)} = \sum_{k=1}^n c_k \mathbf{y}_k(t), \qquad c_k \in \mathbb{K}
+$$
+
+where $\{\mathbf{y_1}, \dots, \mathbf{y_n}\}$ is a linear independent set of solutions, i.e. the basis of the solutions space.
+
+Assume now that $A$ is diagonalizable, and let $\{\mathbf{v_1}, \dots, \mathbf{v_n}\}$ be a basis of $\mathbb{K}^n$ consisting of eigenvectors of A.
+
+$$
+AV = VD, \qquad \text{with } D = \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix}
+$$
+
+then $A = VDV^{-1}$, let $\mathbf{z}(t) = V^{-1} \mathbf{y}(t)$
+
+$$
+\begin{array}{ll}
+&\mathbf{\dot z} = V^{-1} \mathbf{\dot y} = V^{-1} A \mathbf{y} = V^{-1} V D V^{-1} = D \mathbf{z}, \\
+& \mathbf{\dot z} = D \mathbf{z} \implies \mathbf{z}(t) = \mathbf{c} e^{\lambda t}.
+\end{array}
+$$
+
+Obtaining the general solution
+
+$$\mathbf{y}(t) = V \mathbf{z}(t) = \sum_{k=1}^n c_k \mathbf{v_k} e^{\lambda_k t}.
+$$
+
+## Inhomogeneous systems of linear ODEs with constant coefficients
+
+Let $I \subseteq \mathbb{R}$ be an interval, $\mathbf{f}: I \to \mathbb{R}$ continuous. Find functions $\mathbf{y}: I \to \mathbb{R}^n$ such that 
+
+$$
+\mathbf{\dot y}(t) = a \mathbf{y}(t) + \mathbf{f}(t), \qquad t \in I. \qquad (*)
+$$
+
+*Theorem*: let $\mathbf{y}_p: I \to \mathbb{R}^n$ a particular solution for $(*)$ and $\mathbf{y}_H$ the general solution to the homegeneous system. Then the general solutions of the inhomogeneous system $(*)$ is given by
+
+$$
+\mathbf{y}(t) = \mathbf{y}_p(t) + \mathbf{y}_H(t), \qquad t \in I
+$$
+
+*Proof*: similar to 1d case, will possibly be added later.
+
+### Method of variation of parameters
+
+Let $\{\mathbf{y_1}, \dotsc, \mathbf{y_n}\}$ be a basis for the solution space of the homogeneous system. Ansatz:
+
+$$
+\mathbf{y}_p(t) = \sum_{k=1}^n c_k(t) \mathbf{y}_k(t) = (\mathbf{y}_1, \dots, \mathbf{y}_n) \begin{pmatrix} c_1(t) \\ \vdots \\ c_n(t) \end{pmatrix} = Y(t) \mathbf{c}(t),
+$$
+
+where $c_1(t), \dots, c_n(t): I \to \mathbb{R}$ are to be determined. 
+
+Then: 
+
+$$
+\begin{align*}
+\mathbf{\dot y}_p &= \sum_{k=1}^n \dot c_k(t) \mathbf{y}_k(t) + \sum_{k=1}^n c_k(t) \mathbf{\dot y}_k(t), \\
+                  &= \sum_{k=1}^n \dot c_k(t) \mathbf{y}_k(t) + A \sum_{k=1}^n c_k(t) \mathbf{y}_k(t), \\
+                  &= Y(t) \mathbf{\dot c}(t) + A \mathbf{y}_p(t).
+\end{align*}
+$$
+
+Demanding that: $Y(t) \mathbf{\dot c}(t) = \mathbf{f}(t)$. Then $\mathbf{\dot c}(t) = Y^{-1}(t) \mathbf{f}(t) \iff Y(t)$ is nonsingular. Then solve for $\mathbf{c}(t)$.