Laplace Transform

An Integral Transform that converts a function $f(t)$ with $t\in \mathbb{R}$ (usually time domain) to a function $F(s)$ with $s\in \mathbb{C}$ (frequency domain)

$$F(s)=\mathcal{L}\{f(t)\}(s)={\int }_{t=0}^{t=\infty }{e}^{-st}f(t)dt$$

The Laplace Transform is useful because it can convert nth order derivatives into nth order polynomials. Therefore, it can be used to simplify complicated Linear Differential Equations into complicated equations of polynomials.

$$\mathcal{L}\left\{\frac{d^n}{d{x}^n}y(x)\right\}=s\mathcal{L}\left\{\frac{d^{n-1}}{d{x}^{n-1}}y(x)\right\}-\frac{d^{n-1}}{d{x}^{n-1}}y(0)$$

The Laplace transform is an integral transform, so it abides by all the properties of integrals.

$$\mathcal{L}\{Cf(t)\}=C\mathcal{L}\{f(t)\}$$ $$\mathcal{L}\{f(t)+g(t)\}=\mathcal{L}\{f(t)\}+\mathcal{L}\{g(t)\}$$ $$\mathcal{L}\{f(t)\ast g(t)\}=\mathcal{L}\{f(t)\}\cdot \mathcal{L}\{g(t)\}$$

Common Transforms

$f(t)$	$\mathcal{L}\{f(t)\}(s)$
$e^{at}$	$\frac{1}{s-a}$
$\cos (at)$	$\frac{s}{s^2+a^2}$
$\sin (at)$	$\frac{a}{s^2+a^2}$
$t^n{e}^{at}$	$\frac{n!}{(s-a{)}^{n+1}}$
$t^{n}f(t)$	$(-1{)}^n\frac{d^n}{d{s}^n}F(s)$
$H(t-a)f(t-a)$	$e^{-as}F(s)$
$\delta (t-a)$	$e^{-as}$

Example ODE

$$y^{\prime \prime }+y^{\prime }-6y=50e^t,y(0)=c_1,y^{\prime }(0)=c_2$$

Apply Laplace transform to both sides

$$\text{Let }Y(s)=\mathcal{L}\left\{y(t)\right\}(s)$$ $$Y(s)(s^2+s-6)-(s+1)c_1-c_2=\frac{50}{s-1}$$ $$Y(s)=\frac{\frac{50}{s-1}+(s+1)c_1+c_2}{s^2+s-6}=\frac{50+(s-1)(c_{1}s+c_1+c_2)}{(s-1)(s-2)(s+3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s+3}$$ $$\mathcal{L}\{Y(s)\}=\mathcal{L}\{\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s+3}\}$$ $$y(t)=A{e}^t+B{e}^{2t}+C{e}^{-3t}$$

You would do Partial Fractions to get the values of the coefficients here. Though Method of Undetermined Coefficients or Variation of Parameters will be faster in most situations, the Laplace Transform can have its place as a solution method for these kind of problems. For example, the Characteristic Equation for the homogenous equation gives $\lambda =2,-3$, so we instantly get 2/3 of the terms for the solution with minimal effort. Whereas with this method, it takes forever.

ODE with Step Function

$$y^{\prime \prime }+4y^{\prime }+4y=f(t),y(0)=-1,y^{\prime }(0)=2$$ $$f(t)=\{\begin{array}{ll}16e^{2t},& t<3\\ 0,& t\ge 3\end{array}$$ $$f(t)=(1-H(t-3))16e^{2t}=16e^{2t}-H(t-3)\cdot 16e^{2t}$$

where $H$ is the Heaviside Step Function Apply Laplace Transform to both sides.

$$\mathcal{L}\left\{y^{\prime \prime }+4y^{\prime }+4y\right\}=(s^2+4s+4)\cdot Y(s)+s+2$$ $$\mathcal{L}\left\{16e^{2t}-H(t-3)\cdot 16e^{2t}\right\}=\mathcal{L}\left\{16e^{2t}\right\}-\mathcal{L}\left\{H(t-3)\cdot 16e^{2t}\right\}$$ $$\mathcal{L}\left\{16e^{2t}\right\}-\mathcal{L}\left\{H(t-3)\cdot e^6\cdot 16e^{2(t-3)}\right\}=\mathcal{L}\left\{16e^{2t}\right\}-e^6\cdot 16\cdot \mathcal{L}\left\{H(t-3)\cdot g(t-3)\right\}$$ $$\frac{16}{s-2}-16e^6\cdot e^{-3s}G(s)=\frac{16}{s-2}-16e^6\cdot e^{-3s}\frac{1}{s-2}=\frac{16}{s-2}(1-e^{6-3s})$$ $$Y(s)=\left(\frac{16-16e^{6-3s}}{s-2}-s-2\right)\cdot \frac{1}{s^2+4s+4}$$ $$Y(s)=\frac{16-16e^{6-3s}-(s+2)\cdot (s-2)}{(s-2)(s+2{)}^2}=\frac{A}{s-2}+\frac{B}{s+2}+\frac{C}{(s+2{)}^2}$$

Solve the Partial Fractions Take inverse Laplace of both sides.

$$y(t)=A{e}^{2t}+B{e}^{-2t}+Ct{e}^{-2t}$$