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docs/mathematics/ordinary-differential-equations/first-order-ode.md

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+# First-order differential equations
+
+## First-order linear differential equations
+
+A first-order **linear** differential equation is one of the type
+
+$$
+\frac{dy}{dx} + p(x) y = q(x).
+$$
+
+where $p(x)$ and $q(x)$ are given functions, which may be assumed to be continuous. The equation is called **nonhomogeneous** unless $q(x)$ is dentically zero. The corresponding **homogeneous** equation
+
+$$
+\frac{dy}{dx} + p(x)y = 0,
+$$
+
+is separable and so is easily solved. By seperation of variables
+
+$$
+\frac{dy}{dx} = -p(x)y \implies \int \frac{1}{y}dy = \int p(x)dx.
+$$
+
+Though, pay attention to absolute values.
+
+There are two methods for solving nonhomogeneous equations.
+
+### Integration factor
+
+The first method is by using an integrating factor. Let $\mu(x)$ be an antiderivative of $p(x)$ and multiply the equation by $e^{\mu(x)}$.
+
+$$
+\begin{array}{ll}
+e^{\mu(x)} \frac{dy}{dx} + e^{\mu(x)} p(x) y = e^{\mu(x)} q(x) &\implies \frac{d}{dx}(e^{\mu(x)} y) = q(x) e^{\mu(x)}, \\
+&\implies e^{\mu(x)} y = \int q(x) e^{\mu(x)}dx, \\
+&\implies y(x) = e^{-\mu(x)} \int q(x) e^{\mu(x)}dx.
+\end{array}
+$$
+
+### Variation of the constant
+
+The second method is by using a variation of a constant. Let $\mu(x)$ be an antiderivative of $p(x)$ and solve the homegeneous equation
+
+$$
+\frac{dy}{dx} + p(x)y = 0 \implies y(x) = k e^{-\mu(x)}.
+$$
+
+Try:
+
+$$
+y'(x) + p(x) y(x) = q(x) = k'(x) e^{\mu(x)},
+$$
+
+thus $k'(x) = q(x) e^{\mu(x)}$.
+
+#### Example
+
+Solve $\frac{dy}{dx} + 2xy = x$ with $y(0) = 3$.
+
+
+First solving the homogeneous equation
+
+$$
+\begin{array}{ll}
+\frac{dy}{dx} + 2xy = 0 &\implies \int \frac{1}{y} dy = -2\int xdx
+\end{array}
+$$

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

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+# The Laplace transform
+
+*Definition*: let $f: (0,\infty) \to \mathbb{R}$ be a piecewise continuous function that complies to the demand: $\exists s_0 \geq 0, \mu > 0: |f(t)| \leq \mu e^{s_0 t}$, then the **Laplace transform** $\mathcal{L}[f]$ is defined by
+
+$$
+\mathcal{L}[f](s) := \int_0^\infty e^{-st} f(t)dt = F(s),
+$$
+
+where $F(s)$ exists for all $s > s_0$.
+
+## Basic properties
+
+**Linearity**: if $f,g: (0,\infty) \to \mathbb{R}$ both have Laplace transforms, then $f + g$ also has a Laplace transform, and
+
+$$
+\mathcal{L}[f + g] = \mathcal{f} + \mathcal{g},
+$$
+
+on the interval where both are defined. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+If $c \in \mathbb{R}$ then $cf$ also has a Laplace transform and, 
+
+$$
+\mathcal{L}[cf] = c \mathcal{L}[f].
+$$
+
+**Shifting**: if $f$ has a Laplace transform $F$ on $(s_0,\infty)$ and $a \in \mathbb{R}$ then the function $g$ given by
+
+$$
+g(t) = e^{at} f(t)
+$$
+
+has a Laplace transform $G$ on $(\mathrm{max}(s_0 + a),0),\infty$, and
+
+$$
+G(s) = F(s-a)
+$$
+
+on this interval
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+**More shifting**: let $a>0$, if $f$ has a Laplace transform $F$ on $s_0, \infty$ then the function $g$ given by
+
+$$
+g(t) = \begin{cases} f(t-a) \qquad &\text{if } t \geq a, \\ 0 \qquad &\text{if } t < a \end{cases}
+$$
+
+has a Laplace transform G on $(s_0,\infty)$, and
+
+$$
+G(s) = e^{-as}F(s)
+$$
+
+on this interval.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+**Scaling**: let $a > 0$. If $f$ has a Laplace transform $F$ on $(s_0, \infty)$ then the function $g$ given by 
+
+$$
+g(t) = f(at)
+$$
+
+has a Laplace transform G on $(as_0, \infty)$, and 
+
+$$
+G(s) = \frac{1}{a} F\Big(\frac{s}{a}\Big)
+$$
+
+on this interval.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+**Derivatives**: if $f$ has a derivative $g$ having a Laplace transform $G$ on the interval $(s_0,\infty)$ then $f$ has a Laplace transform on the same interval, and
+
+$$
+G(s) = sF(s) - f(0).
+$$
+
+More generally, for higher derivatives we have (under analogous assumptions)
+
+$$
+\mathcal{L}[f^{(n)}](s) = s^n F(s) - \sum_{k=0}^{n-1} s^k f^{(n-1-k)}(0)
+$$
+
+??? note "*Proof*:"
+
+    For large enough $s$, the case $n=1$ follows by integration by parts
+
+    $$
+    \begin{align*} 
+        \mathcal{L}[f'](s) &= \int_0^\infty e^{-st} f'(t)dt, \\ 
+        &= \Big[e^{-st} f(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f(t), \\ 
+        &= sF(s) - f(0),
+    \end{align*}
+    $$
+
+    suppose $\mathcal{L}[f^{k}](s) = s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)$ is true for $k \in \mathbb{N}$. Then by assumption
+
+    $$
+    \begin{align*} 
+        \mathcal{L}[f^{k+1}](s) &= \int_0^\infty e^{-st} f^{(k+1)}(t)dt, \\ 
+        &= \Big[e^{-st} f^{(k+1)}(t) \Big]_0^\infty + s\int_0^\infty e^{-st}f^{(k)}(t), \\ 
+        &= s \mathcal{L}[f^{(k)}] - f^{(k)}(0), \\ &= s \Big(s^k F(s) - \sum_{r=0}^{k-1} s^r f^{(k-1-r)}(0)\Big) - f^{(k)}(0), \\ 
+        &= s^{k+1} F(s) - \sum_{r=0}^{k} s^r f^{(k-r)}(0).
+    \end{align*}
+    $$
+
+## Examples
+
+**Solving a second order linear ODE**: with $y: \mathbb{K} \to \mathbb{R}$ given by 
+
+$$
+\ddot y + 4 \dot y + 4y = t \qquad \text{with } y(0) = 1 \text{ and } \dot y(0) = 0
+$$
+
+using the Laplace transform
+
+$$
+\begin{align*}
+\mathcal{L}[\ddot y + 4 \dot y + 4y](s) &= \frac{1}{s^2} \qquad \text{let } \mathcal{L}[y](s) = Y(s), \\ 
+s^2 Y(s) -s + 4(sY(s) - 1) + 4 Y(s) &= \frac{1}{s^2}, \\ 
+(s^2 + 4s + 4)Y(s) &= \frac{1}{s^2} + s + 4, \\ Y(s) &= \frac{s^3 + 4s^2 +1}{s^2(s+2)^2}, 
+\end{align*}
+$$
+
+then it may be solved with partial fraction decomposition and the inverse transform.
+
+**Solving a linear system of ODEs**: with $\mathbf{y}: \mathbb{K} \to \mathbb{R}^2$ given by
+
+$$
+\mathbf{\dot y}(t) = \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathbf{y}(t) \qquad \text{with } \mathbf{y}(0) = \begin{pmatrix} -3 \\ 7 \end{pmatrix} 
+$$
+
+using the Laplace transform
+
+$$
+\begin{align*} 
+    \mathcal{L}[\mathbf{\dot y}](s) &= \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathcal{L}[\mathbf{y}](s) \qquad \text{let } \mathcal{L}[\mathbf{y}](s) = \mathbf{Y}(s), \\ 
+    s \mathbf{Y}(s) - \mathbf{y}(0) &= \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathbf{Y}(s), \\ 
+    s \mathbf{Y}(s) + \begin{pmatrix} 3 \\ -7 \end{pmatrix} &= \begin{pmatrix} 5 & 1 \\ 1 & 5 \end{pmatrix} \mathbf{Y}(s), \\
+    \begin{pmatrix} 5 - s & 1 \\ 1 & 5 - s \end{pmatrix} \mathbf{Y}(s) &= \begin{pmatrix} 3 \\ -7 \end{pmatrix}, 
+\end{align*}
+$$
+
+using Cramer's rule
+
+$$
+\begin{align*}
+    &Y_1(s) = \frac{\mathrm{det}\begin{pmatrix} 3 & 1 \\ -7 & 5 - s \end{pmatrix}}{(5-s^2)-1}, \\
+    \\
+    &Y_2(s) = \frac{\mathrm{det}\begin{pmatrix} 5 - s & 3 \\ 1 & -7\end{pmatrix}}{(5-s^2)-1},
+\end{align*}
+$$
+
+both can be solved with partial fraction decomposition and the inverse transform.

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

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+# Second-order ordinary differential equations
+
+For simplicity, all definitions and statements are for complex values functions and vector spaces over $\mathbb{C}$.
+
+## Linear second-order ODEs with constant coefficients
+
+Let $L[y] = f$ be given by
+
+$$
+L[y] = \ddot y + p \dot y + qy = f \qquad (*),
+$$
+
+with $f,p,q \in \mathbb{R}$. 
+
+*Definition*: the set of all solutions to $(*)$ is called the general solution.
+
+*Property*: if $y_1,y_2$ are both solutions to the homogeneous case $L[y]=0$ then $\forall c_1,c_2 \in \mathbb{R}$, $y=c_1y_1 + c_2y_2$ is a solution. 
+
+$$
+L[y] = L[c_1y_1 + c_2y_2] = c_1L[y_1] + c_2L[y_2],
+$$
+
+Then the consequence is that the general solution is a linear space.
+
+$(*)$ is said to have **resonance** if $f$ can be split into linearly independent terms of which at least one lies in the solution space of $(*)$.
+
+### Solving homogeneous linear second-order ODEs with constant coefficients
+
+Therefore solving
+
+$$
+L[y] = \ddot y + p \dot y + qy = 0.
+$$
+
+Ansatz: let $y(t) = e^{\lambda t}$ with $\lambda \in \mathbb{C}$. Then
+
+$$
+L[y(t)] = \lambda^2 e^{\lambda t} + p \lambda e^{\lambda t} + q e^{\lambda t} = e^{\lambda t} (\lambda^2 + p \lambda + q) = 0,
+$$
+
+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 
+
+$$
+y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}, \quad c_1,c_2 \in \mathbb{C},
+$$
+
+if instead one root $\lambda_1 \in \mathbb{C}$ is foundt the solution space is
+
+$$
+y(t) = (c_1 + c_2t) e^{\lambda_1 t}.
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+#### 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
+
+$$
+y(t) = c_1 e^{(-2 + 2i)1 t} + c_2 e^{(-2 - 2i) t}, \quad c_1,c_2 \in \mathbb{C},
+$$
+
+and we can write the real solution as
+
+$$
+y(t) = e^{-2t}\big(d_1\cos 2t + d_2 \sin 2t \big), \quad d_1,d_2 \in \mathbb{R}.
+$$
+
+### Solving inhomogeneous linear second-order ODEs with constant coefficients
+
+*Theorem*: let $y_p$ be a particular solution to $(*)$. Then the general solution to $(*)$ is given by
+
+$$
+y = y_h + y_p,
+$$
+
+with $y_h$ the solution to the homegeneous case.
+
+??? note "*Proof*:"
+
+    Let $y$ be a solution to $(*)$, then $L[y - y_p] = L[y] - L[y_p] = f - f = 0$. Therefore $y = (y - y_p) + y_p = y_h + y_p$.
+
+#### Method of variation of parameters
+
+We need the general solution to the homogeneous case 
+
+$$
+y_h(t) = c_1 y_1(t) + c_2 y_2(t), \qquad c_1,c_2 \in \mathbb{C}. 
+$$
+
+Ansatz: let $y_p(t) = c_1(t) y_2(t) + c_2(t) y_2(t)$, then taking the derivative of $y_p(t)$
+
+$$
+\dot y_p(t) = \dot c_1(t) y_2(t) + \dot c_2(t) y_2(t) + c_1(t) \dot y_2(t) + c_2(t) \dot y_2(t),
+$$
+
+we demand that $\dot c_1(t) y_2(t) + \dot c_2(t) y_2(t) = 0$. Then taking the second derivative of $y_p(t)$
+
+$$
+\ddot y_p(t) = \dot c_1(t) \dot y_2(t) + \dot c_2(t) \dot y_2(t) + c_1(t) \ddot y_2(t) + c_2(t) \ddot y_2(t),
+$$
+
+then we have for $(*)$
+
+$$
+\ddot y_p(t) + p \dot y_p(t) + q = c_1\big(\ddot y_1 + p \dot y_1 + q y_1\big) + c_2\big(\ddot y_2 + p \dot y_2 + q y_2\big) + \dot c_1 \dot y_1 + \dot c_2 \dot y_2 = f
+$$
+
+we demand that $\dot c_1 \dot y_1 + \dot c_2 \dot y_2 = f$. Then we can create a linear system of demands
+
+$$
+\begin{pmatrix} y_1 && y_2 \\ \dot y_1 && \dot y_2\end{pmatrix} \begin{pmatrix} \dot c_1 \\ \dot c_2 \end{pmatrix} = \begin{pmatrix} 0 \\ f \end{pmatrix},
+$$
+
+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 $\alpha \in \mathbb{R}$
+
+| Inhomogeneity | Particular solution |
+| ------ | --------------- |
+| $L[y] = P_n$ | $t^m A_n$ |
+| $L[y] = P_n e^{\alpha t}$ | $t^m A_n e^{\alpha t}$ |
+| $L[y] = P_n \cos \omega t$ | $t^m \big(A_n \cos \omega t + B_n \sin \omega t \big)$ |
+| $L[y] = P_n \sin \omega t$ | $t^m \big(A_n \cos \omega t + B_n \sin \omega t \big)$ |
+| $L[y] = P_n e^{\alpha t} \cos \omega t$ | $t^m e^{\alpha t} \big(A_n \cos \omega t + B_n \sin \omega t \big)$ |
+| $L[y] = P_n e^{\alpha t} \sin \omega t$ | $t^m e^{\alpha t} \big(A_n \cos \omega t + B_n \sin \omega t \big)$ |
+
+Choose $m \in \mathbb{N} \cup \{0\}$ as small as possible such that no term in the ansatz solves the homogeneous equation $L[y] = 0$.
+

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

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+# 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,
+
+$$
+\mathbf{y}(t) = \sum_{k=1}^n c_k \mathbf{y}_k(t), \qquad c_k \in \mathbb{K}
+$$
+
+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
+
+$$
+\mathbf{y}(t) = \mathbf{y}_p(t) + \mathbf{y}_h(t), \qquad t \in I
+$$
+
+??? note "*Proof*:"
+
+    Similar to 1d case, will 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: $Y(t) \mathbf{\dot c}(t) = \mathbf{f}(t)$ is the Wronskian. Then $\mathbf{\dot c}(t) = Y^{-1}(t) \mathbf{f}(t) \iff Y(t)$ is nonsingular. Then solve for $\mathbf{c}(t)$.