MATH1012 - Lecture 30

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MATH1012 - Lecture 29
Skeleton Slides

Linearity of the Laplace Transformation

• If $F(s)=\mathcal{L}(f)(s)$ and $G(s)=\mathcal{L}(g)(s)$ then: $\mathcal{L}(\alpha f+\beta g)(s)=\alpha \mathcal{L}(f)(s)+\beta \mathcal{L}(g)(s)$
  • Constants $\alpha ,\beta \in \mathbb{R}$
• If we can split up a function into parts, we make it easier to solve
  • We are given set Laplace transformations for many different functions already (see formula sheet)
  • May need to use techniques such as partial fractions
• This applies to inverse Laplace transformations as well

Laplace Transform of the Derivative

• If $f(t)$ is differentiable and has Laplace transform $F(s)$ then
  $\mathcal{L}(f^{\prime })=sF(s)-f(0)$
• This process can be seen again for $\mathcal{L}(f^{\prime \prime })$:
  $\mathcal{L}(f^{\prime \prime })=s^{2}F(s)-sf(0)-f^{\prime }(0)$
  General formula: $\mathcal{L}(f^{(n)})=s^{n}F(s)-s^{n-1}f(0)-\cdots -s{f}^{(n-2)}(0)-f^{(n-1)}(0)$
• We see that $\mathcal{L}(f^{(n)})$ involves no derivatives of $F(s)$
• Its simply a multiple $s^n$ of $F(s)$ plus a polynomial of degree $n-1$ in $s$

Laplace Transformation of a Definite Integral

$\mathcal{L}\left({\int }_0^{t}f(u)du\right)=\frac{F(s)}{s}$

• This can be helpful to find inverse transforms of functions which have a factor $s$ appear in the denominator
  • Example in lecture recording (15th May)
• i.e. use to solve when asked to find ${\mathcal{L}}^{-1}(\frac{F(s)}{s})$

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MATH1012 - Lecture 31
Annotated Slides