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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 \(\lim_{c \to \infty} \int_0^c e^{-st}f(t)\;dt\). 


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


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




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


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 \(y(0)=1\)


\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 \(y(0)=0, y'(0)=4\)


\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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