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Laplace Transforms - Definition and Properties

You will learn what the Laplace transform is, how to compute it for common functions, and understand its key properties, especially how it turns differentiation into multiplication by ss.

The Laplace transform of a function f(t)f(t) is defined as:

L{f(t)}=F(s)=0estf(t)dt\mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt

It converts a function of time tt into a function of a new variable ss. The real power comes when we take the Laplace transform of a differential equation:

L{y}=s2Y(s)sy(0)y(0)\mathcal{L}\{y''\} = s^2 Y(s) - s\,y(0) - y'(0) L{y}=sY(s)y(0)\mathcal{L}\{y'\} = s\,Y(s) - y(0) L{y}=Y(s)\mathcal{L}\{y\} = Y(s)

This turns a differential equation into a purely algebraic equation that is often much easier to solve. We then use the inverse Laplace transform to get back to the time domain.

The diagram shows the Laplace transform workflow: start with a differential equation in the time domain, transform to the s-domain where it becomes algebra, solve for Y(s), then inverse-transform back to get y(t).

L{f(t)}=sF(s)f(0)\mathcal{L}\{f'(t)\} = s\,F(s) - f(0) L{f(t)}=s2F(s)sf(0)f(0)\mathcal{L}\{f''(t)\} = s^2 F(s) - s\,f(0) - f'(0) L{eat}=1sa\mathcal{L}\{e^{at}\} = \frac{1}{s - a} L{sinbt}=bs2+b2,L{cosbt}=ss2+b2\mathcal{L}\{\sin bt\} = \frac{b}{s^2 + b^2}, \quad \mathcal{L}\{\cos bt\} = \frac{s}{s^2 + b^2} L{tn}=n!sn+1\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}

Example 1: Basic Transform

Find L{t2+3sin2t}\mathcal{L}\{t^2 + 3\sin 2t\}.

Solution:

L{t2}+3L{sin2t}=2s3+32s2+4=2s3+6s2+4\mathcal{L}\{t^2\} + 3\mathcal{L}\{\sin 2t\} = \frac{2}{s^3} + 3 \cdot \frac{2}{s^2 + 4} = \frac{2}{s^3} + \frac{6}{s^2 + 4}

Example 2: Solving a Simple IVP

Solve y+3y=0y' + 3y = 0 with y(0)=5y(0) = 5 using Laplace transforms.

Solution: Take Laplace of both sides:

sY5+3Y=0    Y(s)=5s+3sY - 5 + 3Y = 0 \implies Y(s) = \frac{5}{s + 3}

Inverse: y(t)=5e3ty(t) = 5e^{-3t}

Example 3: Second-Order IVP

Solve y+4y=0y'' + 4y = 0 with y(0)=1y(0) = 1, y(0)=0y'(0) = 0.

Solution:

s2Ys+4Y=0    Y(s)=ss2+4s^2 Y - s + 4Y = 0 \implies Y(s) = \frac{s}{s^2 + 4}

Inverse: y(t)=cos2ty(t) = \cos 2t

Laplace transforms are essential in control theory, electrical engineering (analyzing circuits with sudden switches or AC inputs), mechanical engineering (vibration analysis), and signal processing. They make it dramatically easier to solve initial value problems, especially when there are discontinuous forcing functions like step inputs or impulses.

The Laplace transform converts:
$\mathcal{L}\{e^{at}\}$ equals:
The main advantage of Laplace transforms for IVPs is:
$\mathcal{L}\{\sin bt\}$ equals:
Laplace transforms are particularly powerful when dealing with: