MATH1012 - Lecture 31

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MATH1012 - Lecture 30
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Using Laplace Transforms to Solve DEs

For MATH1012, we will limit ourselves to considering 2nd order DEs with constant coefficients and two initial conditions

• Given 2nd order DE: $\frac{d^{2}y}{d{t}^2}+p\frac{dy}{dt}+qy=g(t)$, applying Laplace transforms:
  $s^{2}Y(s)-s{y}_0-v_0+p(sY(s)-y_0)+qY(s)=G(s)$
  Where $Y(s)=\mathcal{L}(y)$ and $G(s)=\mathcal{L}(g)$
  $y(0)=y_0$ and $\frac{dy}{dx}(0)=v_0$
  $g(t),y_0$ and $v_0$ are given
• We then solve for $Y(s)$ and invert this to get $y(t)$
  • This method will often require the use of partial fractions
  • It is much easier to use the finger-method while solving these

This method is usually not preferred over the other for most 2nd order DEs. However, for Heaviside functions (see next lecture), we need to use Laplace transforms as the other method will not work.

The Derivative of a Laplace Transform

• Suppose $F(s)$ is the Laplace transform of $f(t)$, then:
  $\mathcal{L}(tf(t))=-F^{\prime }(s)$
  $F^{\prime }(s)$ is the derivative of $F(s)$
• e.g. the Laplace transform of $t{e}^{at}$ [$\mathcal{L}(e^{at})=\frac{1}{s-a}$, given by formula sheet]
  • $\mathcal{L}\{t{e}^{at}\}=-\frac{d}{ds}(\frac{1}{s-a})=\frac{1}{(s-a{)}^2}$

The S-Shift Theorem

• If $F(s)$ is the Laplace transform of $f(t)$, then:
  $\mathcal{L}(e^{at}f(t))=F(s-a)$
  Proof in lecture recording
• Given ${\mathcal{L}}^{-1}(G(s))$, try rearrange $G(s)$ into a form $F(s-a)$ to solve
  • Where $F(s)$ is a function from our formula sheet

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MATH1012 - Lecture 32
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