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 # ECE 205: Advanced Calculus 1
 
+## Laplace transform
+
+The Laplace transform is a wonderful operation to convert a function of $t$ into a function of $s$. Where $s$ is an unknown variable independent of $t$:
+
+$$
+\mathcal L\{f(t)\}=F(s)=\int^\infty_0e^{-st}f(t)dt, s > 0
+$$
+
+??? example
+    To solve for $\mathcal L\{\sin(at)\}$:
+
+    \begin{align*}
+    \mathcal L\{f(t)\}&=\int^\infty_0e^{-st}\sin(at)dt \\
+    \\
+    \text{IBP: let $u=\sin(at)$, $dv=e^{-st}dt$:} \\
+    &=\lim_{B\to\infty} \underbrace{\biggr[
+    	\cancel{-\frac 1 se^{-st}\sin(at)}}_\text{0 when $s=0$ or $s=\infty$}+\frac a s\int e^{-st}\cos(at)dt
+    \biggr]^B_0 \\
+    &=\frac a s\lim_{B\to\infty}\left[\int e^{-st}\cos(at)dt \right]^B_0 \\
+    \text{IBP: let $u=\cos(at)$, $dv=e^{-st}dt$:} \\
+    &=\frac a s \lim_{B\to\infty}\left[
+    	-\frac 1 s e^{-st}\cos(at)-\frac a s\underbrace{\int e^{-st}\sin(at)dt}_{\mathcal L\{\sin(at)\}}
+    \right]^B_0 \\
+    &=\frac{a}{s^2}-\frac{a^2}{s^2}\mathcal L\{\sin(at)\} \\
+    \mathcal L\{\sin(at)\}\left(1+\frac{a^2}{s^2}\right)&=\frac{a}{s^2} \\
+    \mathcal L\{\sin(at)\}&=\frac{a}{a^2+s^2}, s > 0
+    \end{align*}
+
+A **piecewise continuous** function on $[a,b]$ is continuous on $[a,b]$ except for a possible finite number of finite jump discontinuities.
+
+- This means that any jump discontinuities must have a finite limit on both sides.
+- A piecewise continuous function on $[0,\infty)$ must be piecewise continuous $\forall B>0, [0,B]$.
+   
+The **exponential order** of a function is $a$ if there exist constants $K, M$ such that:
+$$|f(t)|\leq Ke^{at}\text{ when } t\geq M$$
+
+!!! example
+    - $f(t)=7e^t\sin t$ has an exponential order of 1.
+    - $f(t)=e^{t^2}$ does not have an exponential order.
+
+### Linearity
+
+A **piecewise continuous** function $f$ on $[0,\infty)$ of an exponential order $a$ has a defined Laplace transform for $s>a$.
+
+Laplace transforms are **linear**. If there exist LTs for $f_1, f_2$ for $s>a_1, a_2$, respectively, for $s=\text{max}(a_1, a_2)$:
+
+$$\mathcal L\{c_1f_1 + c_2f_2\} = c_1\mathcal L\{f_1\} + c_2\mathcal L\{f_2\}$$
+
+??? example
+    We find the Laplace transform for the following.
+    
+    $$
+    f(t)=\begin{cases}
+    1 & 0\leq t < 1 \\
+    e^{-t} & t\geq 1
+    \end{cases}
+    $$
+    
+    Clearly $f(t)$ is piecewise ocontinuous on $[0,\infty)$ and has an exponential order of -1 when $t\geq 1$ and 0 when $0\leq t<1$. Thus $\mathcal L\{f(t)\}$ is defined for $s>0$.
+    
+    \begin{align*}
+    \mathcal L\{f(t)\}&=\int^1_0 e^{-st}dt + \int^\infty_1e^{-st}e^{-t}dt \\
+    \tag{$s\neq 0$}&=\left[-\frac 1 s e^{-st}\right]^1_0 + \int^\infty_1e^{t(-s-1)}dt \\
+    &=-\frac 1 se^{-s}+\frac 1 s + \lim_{B\to\infty}\left[ \frac{1}{-s-1}e^{t(-s-1)} \right]^B_1 \\
+    \tag{$s\neq 0,s>-1$}&=\frac{-e^{-s}+1}{s} -\frac{e^{-s-1}}{-s-1}
+    \end{align*}
+    
+    We solve for the special case $s=0$:
+    \begin{align*}
+    \mathcal L\{f(t)\}&=\int^1_0 e^{0}dt + \int^\infty_1e^{-st}e^{-t}dt \\
+    &=1 -\frac{e^{-s-1}}{-s-1} \\
+    \end{align*}
+    
+    $$
+    \mathcal L\{f(t)\}=
+    \begin{cases}
+    \frac{-e^{-s}+1}{s}-\frac{e^{-s-1}}{-s-1} & s\neq 0, s>-1 \\
+    1-\frac{e^{-s-1}}{-s-1} &s=0
+    \end{cases}
+    $$
+
+If there exists a transform for $s>a$, the original function multiplied by $e^{-bt}$ exists for $s>a+b$.
+
+$$\mathcal L\{f(t)\}=F(s), s>a\implies \mathcal L\{e^{-bt}f(t)\}=F(s),s>a+b$$
+
+### Inverse transform
+
+The inverse is found by manipulating the equation until you can look it up in the [Laplace Table](#resources).
+
+The inverse transform is also **linear**.
+
+### Inverse of rational polynomials
+
+If the transformed function can be expressed as a partial fraction decomposition, it is often easier to use linearity to reference the table.
+
+$$\mathcal L^{-1}\left\{\frac{P(s)}{Q(s)}\right\}$$
+
+- $Q, P$ are polynomials
+- $\text{deg}(P) > \text{deg}(Q)$
+- $Q$ is factored
+
+??? example
+    \begin{align*}
+    \mathcal L^{-1}\left\{\frac{s^2+9s+2}{(s-1)(s^2+2s-3)}\right\} &=\mathcal L^{-1}\left\{\frac{A}{s-1}+\frac{B}{s+3} + \frac{Cs+D}{(s-1)^2}\right\} \\
+    &\implies A=2,B=3,C=-1 \\
+    &=2\mathcal L^{-1}\left\{\frac{1}{s-1}\right\} + 3\mathcal L^{-1}\left\{\frac{1}{(s-1)^2}\right\}-\mathcal L^{-1}\left\{\frac{1}{s+3}\right\} \\
+    &=2e^t+3te^t-e^{-3t}
+    \end{align*}
+
+### Inverse of differentiable equations
+
+If a function $f$ is continuous on $[0,\infty)$ and its derivative $f'$ is piecewise continuous on $[0,\infty)$, for $s>a$:
+
+$$
+\mathcal L\{ f'\}=s\mathcal L\{f\}-f(0) \\
+\mathcal L\{ f''\} = s^2\mathcal L\{f\}-s\cdot f(0)-f'(0)
+$$
+
+### Solving IVPs
+
+Applying the Laplace transform to both sides of an IVP is valid to remove any traces of horrifying integration.
+
+!!! example
+    \begin{align*}
+    y''-y'-2y=0, y(0)=1, y'(0)=0 \\
+    \mathcal L\{y''-y'-2y\}&=\mathcal L\{0\} \\
+    s^2\mathcal L\{y\}-s\cdot y(0)-y'(0) - s\mathcal L\{y\} +y(0) - 2\mathcal L\{y\}&=0 \\
+    \mathcal L\{y\}(s^2-s-2)-s+1&=0 \\
+    \mathcal L\{y\}&=\frac{s-1}{(s-2)(s+1)} \\
+    &= \\
+    \mathcal L^{-1}\{\mathcal L\{y\}\}&=\mathcal L^{-1}\left\{
+   	 \frac 1 3\cdot\frac{1}{s-2} + \frac 2 3\cdot\frac{1}{s+1}
+    \right\} \\
+    y&=\frac 1 3\mathcal L^{-1}\left\{\frac{1}{s-2}\right\} + \frac 2 3\mathcal L^{-1}\left\{\frac{1}{s+1}\right\} \\
+    \tag{from Laplace table}&=\frac 1 3 e^{2t} + \frac 2 3 e^{-t}
+    \end{align*}
+
+### Heaviside / unit step
+
+The Heaviside and unit step functions are identical:
+
+$$
+H(t-c)=u(t-c)=u_c(t)=\begin{cases}
+0 & t < c \\
+1 & t \geq c
+\end{cases}
+$$
+
+Piecewise continuous functions can be manipulated into a single equation via the Heaviside function.
+
+For a Heaviside transform $\mathcal L\{u_c(t)g(t)\}$, if $g$ is defined on $[0,\infty)$, $c\geq 0$, and $\mathcal L\{g(t+c)\}$ exists for some $s>s_0$:
+
+$$
+\mathcal L\{u_c(t)g(t)\}=e^{-sc}\mathcal L\{g(t+c)\},s>s_0
+$$
+
+## Resources
+
+- [Laplace Table](/resources/ece/laplace.pdf)