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Calculus

Definition

LaPlace transforms give us an algebraic method to solve differential equations when given an initial condition. Recall that initial conditions are restrictions on a differential equation that give us a particular solution to a differential equation.

The official definition of a LaPlace Transform, given a function $$f(t)$$, is

$F(s) = \mathscr{L}(f) = \int_0^{\infty} e^{-st}f(t)\;dt$

Note that since the upper limit is $$\infty$$, this is an improper integral. This is the same as evaluating $$\lim_{c \to \infty} \int_0^c e^{-st}f(t)\;dt$$.

Example

Find the Laplace transform of $$f(t)=1$$.

Solution

Using the definition, we get

\begin{align} \mathscr{L}(f) &= \int_0^{\infty} e^{-st}f(t)\;dt \\ \mathscr{L}(1) &= \int_0^{\infty} e^{-st}(1)\;dt \\ &= \lim_{c \to \infty} \int_0^c e^{-st}(1)\;dt \\ &= \lim_{c \to \infty} \int_0^c e^{-st}\;dt \\ &= \lim_{c \to \infty} \frac{-1}{s}e^{-st}|_0^c \\ &= \lim_{c \to \infty}\frac{-1}{s}(e^{-st}-1) \\ &= \frac{-1}{s}(-1) \\ &= \frac{1}{s} \end{align}
Thus,  $$\mathscr{L}(f) = \frac{1}{s}$$

Table of Laplace Transformations

 $$f(t)=\mathscr{L}^{-1}(F)$$ $$\mathscr{L}(f)=F(s)$$ $$f(t)=\mathscr{L}^{-1}(F)$$ $$\mathscr{L}(f)=F(s)$$ $$1$$ $$\frac{1}{s}$$ $$\frac{t^{n-1}}{(n-1)!}$$ $$\frac{1}{s^{n}}(n=1,2,3,...)$$ $$e^{-at}$$ $$\frac{1}{s+a}$$ $$1-e^{-at}$$ $$\frac{a}{s(s+a)}$$ $$\cos{at}$$ $$\frac{s}{s^2+a^2}$$ $$\sin{at}$$ $$\frac{s}{s^2+a^2}$$ $$1-\cos{at}$$ $$\frac{a^2}{s(s^2+a^2)}$$ $$at-\sin{at}$$ $$\frac{a^3}{s(s^2+a^2)}$$ $$e^{-at}-e^{-bt}$$ $$\frac{b-a}{(s+a)(s+b)}$$ $$ae^{-at}-ae^{-bt}$$ $$\frac{s(a-b)}{(s+a)(s+b)}$$ $$te^{-at}$$ $$\frac{1}{(s+a)^2}$$ $$t^{n-1}e^{-at}$$ $$\frac{(n-1)!}{(s+a)^n}$$ $$e^{-at}(1-at)$$ $$\frac{s}{(s+a)^2}$$ $$[(b-a)t+1]e^{-at}$$ $$\frac{s+b}{(s+a)^2}$$ $$\sin{at}-at\cos{at}$$ $$\frac{2a^3}{(s^2+a^2)^2}$$ $$t\sin{at}$$ $$\frac{2as}{(s^2+a^2)^2}$$ $$\sin{at}+at\cos{at}$$ $$\frac{2as^2}{(s^2+a^2)^2}$$ $$t\cos{at}$$ $$\frac{s^2-a^2}{(s^2+a^2)^2}$$ $$e^{-at}\sin{bt}$$ $$\frac{b}{(s+a)^2+b^2}$$ $$e^{-at}\cos{bt}$$ $$\frac{s+a}{(s+a)^2+b^2}$$

Properties of Laplace Transforms

Below are some properties of Laplace transforms. These properties will become useful when solving differential equations using Laplace transforms.

1. Linearity Property: $$\mathscr{L}[af(t)+bg(t)]=a\mathscr{L}(f)+b\mathscr{L}(g)$$
2. Laplace transform of a function's first derivative: $$\mathscr{L}(f')=s\mathscr{L}(f)-f(0)$$
3. Laplace transform of a function's second derivative: $$\mathscr{L}(f'')=s^2\mathscr{L}(f)-sf(0)-f'(0)$$

We also have inverse transforms which, given a Laplace transform, will give us the corresponding function

Notation: $$\mathscr{L}^{-1}(F)=f(t)$$ where $$\mathscr{L}^{-1} denotes the inverse transform Solving Differential Equations Using the above information, we can now solve differential equations, given an initial condition. Let's look at some examples. Example 1 Solve this first order differential equation: \(y'-3y=0$$ given $$y(0)=1$$

Solution

\begin{align} \mathscr{L}(y'-3y) &= \mathscr{L}(0) & \text{take transform of both sides} \\ \mathscr{L}(y')-3\mathscr{L}(y) &= \mathscr{L}(0) & \text{by property 1 (linearity)} \\ s\mathscr{L}(y)-y(0)-3\mathscr{L}(y) &= \mathscr{L}(0) & \text{by property 2} \\ s\mathscr{L}(y)-1-3\mathscr{L}(y) &= 0 & \text{by definition } (y(0)=1, \mathscr{L}(0)=0) \\ s\mathscr{L}(y)-3\mathscr{L}(y) &= 1 & \\ \mathscr{L}(y)(s-3) &= 1 & \\ \mathscr{L}(y) &= \frac{1}{s-3} & \\ y(t) &= e^{3t} & \text{take inverse transform} \end{align}

Thus, the solution is $$y=e^{3t}$$. This can be verified by differentiating $$y$$ and substituting it back into the original equation.

Example 2
Solve this second-order differential equation: $$2y''+8y'=0$$ given $$y(0)=0, y'(0)=4$$

Solution

\begin{align} \mathscr{L}(2y''+8y') &= \mathscr{L}(0) & \text{take transform of both sides} \\ 2\mathscr{L}(y'') +8\mathscr{L}(y') &= \mathscr{L}(0) & \text{by proposition 1} \\ 2[s^2\mathscr{L}(y)-sy(0)-y'(0)]+8[s\mathscr{L}(y)-y(0)] &= \mathscr{L}(0) & \text{by properties 2 and 3} \\ 2[s^2\mathscr{L}(y)-sy(0)-4]+8[s\mathscr{L}(y)-0] &= 0 & \text{substituting values} \\ 2s^2\mathscr{L}(y)-8+8s\mathscr{L}(y) &= 0 & \\ 2s^2\mathscr{L}(y) + 8s\mathscr{L} &= 8 & \\ \mathscr{L}(y)[2s^2+8s] &= 8 & \\ \mathscr{L}(y) &= \frac{8}{2s^2+8s} & \\ \mathscr{L}(y) &= \frac{8}{2s(s+4)} & \\ \mathscr{L}(y) &= \frac{4}{s(s+4)} & \\ y(t) &= 1-e^{-4t} & \text{take inverse transform} \end{align}

We can check the validity of this solution by finding the derivative and substituting it back into our original equation.

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