feat: add laplace equations

Author: SuhasDissa

pnpm-workspace.yaml

@@ -0,0 +1,119 @@
+---
+title: "1. Complete Laplace Transform Table"
+description: ""
+---
+
+## Basic Properties and Theorems
+
+### Linearity
+$$\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$$
+
+### First Shifting Theorem (Frequency Shift)
+$$\mathcal{L}\{e^{at}f(t)\} = F(s-a)$$
+
+### Second Shifting Theorem (Time Shift)
+$$\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s), \quad a > 0$$
+
+### Scaling Theorem
+$$\mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right), \quad a > 0$$
+
+### Differentiation (Time Domain)
+$$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$$
+$$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$$
+
+### Integration (Time Domain)
+$$\mathcal{L}\left\{\int_0^t f(\tau)d\tau\right\} = \frac{F(s)}{s}$$
+
+### Differentiation (Frequency Domain)
+$$\mathcal{L}\{tf(t)\} = -\frac{dF}{ds}$$
+$$\mathcal{L}\{t^nf(t)\} = (-1)^n\frac{d^nF}{ds^n}$$
+
+### Convolution Theorem
+$$\mathcal{L}\{f(t) * g(t)\} = F(s)G(s)$$
+where $f(t) * g(t) = \int_0^t f(\tau)g(t-\tau)d\tau$
+
+### Initial Value Theorem
+$$f(0^+) = \lim_{s \to \infty} sF(s)$$
+
+### Final Value Theorem
+$$\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$$ (if limit exists)
+
+---
+
+## Standard Laplace Transform Pairs
+
+| $f(t)$ | $F(s) = \mathcal{L}\{f(t)\}$ | Conditions |
+|--------|-----|-----------|
+| $\delta(t)$ (Dirac delta) | $1$ | |
+| $u(t)$ (Unit step) | $\frac{1}{s}$ | $s > 0$ |
+| $t$ | $\frac{1}{s^2}$ | $s > 0$ |
+| $t^n$ | $\frac{n!}{s^{n+1}}$ | $s > 0, n \in \mathbb{Z}^+$ |
+| $t^{n-1}e^{-at}$ | $\frac{(n-1)!}{(s+a)^n}$ | $s > -a, n \in \mathbb{Z}^+$ |
+| $e^{at}$ | $\frac{1}{s-a}$ | $s > a$ |
+| $te^{at}$ | $\frac{1}{(s-a)^2}$ | $s > a$ |
+| $t^ne^{at}$ | $\frac{n!}{(s-a)^{n+1}}$ | $s > a$ |
+| $\sin(bt)$ | $\frac{b}{s^2+b^2}$ | $s > 0$ |
+| $\cos(bt)$ | $\frac{s}{s^2+b^2}$ | $s > 0$ |
+| $\sinh(bt)$ | $\frac{b}{s^2-b^2}$ | $s > b$ |
+| $\cosh(bt)$ | $\frac{s}{s^2-b^2}$ | $s > b$ |
+| $e^{at}\sin(bt)$ | $\frac{b}{(s-a)^2+b^2}$ | $s > a$ |
+| $e^{at}\cos(bt)$ | $\frac{s-a}{(s-a)^2+b^2}$ | $s > a$ |
+| $t\sin(bt)$ | $\frac{2bs}{(s^2+b^2)^2}$ | $s > 0$ |
+| $t\cos(bt)$ | $\frac{s^2-b^2}{(s^2+b^2)^2}$ | $s > 0$ |
+| $\sin(bt) - bt\cos(bt)$ | $\frac{2b^3}{(s^2+b^2)^2}$ | $s > 0$ |
+| $t\sinh(bt)$ | $\frac{2bs}{(s^2-b^2)^2}$ | $s > b$ |
+| $t\cosh(bt)$ | $\frac{s^2+b^2}{(s^2-b^2)^2}$ | $s > b$ |
+
+---
+
+## Additional Common Functions
+
+| $f(t)$ | $F(s)$ |
+|--------|--------|
+| $u(t-a)$ | $\frac{e^{-as}}{s}$ |
+| $(t-a)u(t-a)$ | $\frac{e^{-as}}{s^2}$ |
+| $e^{-at}u(t)$ | $\frac{1}{s+a}$ |
+| $\sin(bt-c)u(t)$ | $\frac{b\cos(c) + s\sin(c)}{s^2+b^2}$ |
+| $\cos(bt-c)u(t)$ | $\frac{s\cos(c) + b\sin(c)}{s^2+b^2}$ |
+| $\frac{\sin(bt)}{t}$ | $\arctan\left(\frac{b}{s}\right)$ |
+| $u(t) - u(t-a)$ (pulse) | $\frac{1-e^{-as}}{s}$ |
+| $e^{at} - e^{bt}$ | $\frac{1}{s-a} - \frac{1}{s-b}$ |
+
+---
+
+## Partial Fraction Decomposition (for Inverse Transforms)
+
+When inverting rational functions, use partial fractions:
+
+### Case 1: Distinct Linear Factors
+$$\frac{P(s)}{(s-a)(s-b)} = \frac{A}{s-a} + \frac{B}{s-b}$$
+
+### Case 2: Repeated Linear Factors
+$$\frac{P(s)}{(s-a)^n} = \frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + \cdots + \frac{A_n}{(s-a)^n}$$
+
+### Case 3: Quadratic Factors
+$$\frac{P(s)}{(s^2+bs+c)} = \frac{As+B}{s^2+bs+c}$$
+
+Complete the square: $s^2+bs+c = (s+\frac{b}{2})^2 + (c-\frac{b^2}{4})$
+
+---
+
+## Common Inverse Transform Patterns
+
+### Pattern 1: $\frac{1}{s^2+b^2}$
+$$\mathcal{L}^{-1}\left\{\frac{1}{s^2+b^2}\right\} = \frac{\sin(bt)}{b}$$
+
+### Pattern 2: $\frac{s}{s^2+b^2}$
+$$\mathcal{L}^{-1}\left\{\frac{s}{s^2+b^2}\right\} = \cos(bt)$$
+
+### Pattern 3: $\frac{1}{(s-a)^2+b^2}$
+$$\mathcal{L}^{-1}\left\{\frac{1}{(s-a)^2+b^2}\right\} = \frac{e^{at}\sin(bt)}{b}$$
+
+### Pattern 4: $\frac{s-a}{(s-a)^2+b^2}$
+$$\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2+b^2}\right\} = e^{at}\cos(bt)$$
+
+### Pattern 5: $\frac{1}{s(s+a)}$
+$$\mathcal{L}^{-1}\left\{\frac{1}{s(s+a)}\right\} = \frac{1}{a}(1-e^{-at})$$
+
+### Pattern 6: $\frac{1}{(s-a)(s-b)}, a \neq b$
+$$\mathcal{L}^{-1}\left\{\frac{1}{(s-a)(s-b)}\right\} = \frac{e^{at}-e^{bt}}{a-b}$$