Laplace Transforms Tutorial Sheet, Sheet #9

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Skill Building Questions

Problem 1.

Using the given definition, derive the following laplace transforms:

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

(a) $f(t)=e^{at}$

$\Rightarrow{}\quad$ $F(s)=\int_{0}^{\infty}{e^{-st}{e^{at}}}{dt}=\int_{0}^{\infty}{e^{-(s-a)t}{dt}}=\left[-\frac{1}{s-a}e^{-(s-a)t}\right]_0^\infty, \quad{s>a}$
$\Rightarrow{}\quad\boxed{F(s)=\frac{1}{s-a}}$








(b) $g(t)=9$

$\Rightarrow{}\quad$ $G(s)=\int_{0}^{\infty} \ {9e^{-st}{dt}}={[-\frac{9}{s}{e^{-st}}]_0^\infty}\quad$
$\Rightarrow{}\quad\boxed{G(s)=\frac{9}{s}}$








(c) $k(t)=4t$

$\Rightarrow{}\quad$ $K(s)=\int_{0}^{\infty} 4t \ e^{-st} dt=4\int_{0}^{\infty} t \ e^{-st} dt$
Using integration by parts: $udv=uv-\int{vdu}$
$u=t, dv=e^{-st}dt$ and $du=dt,$
$v=-\frac{1}{s} e^{-st}$
$\Rightarrow{}\quad$ $4\int_{0}^{\infty} t \ e^{-st}dt=4(-\frac{t}{s} e^{-st} - \int_{0}^{\infty} \frac{-1}{s} e^{-st}dt)=4(\frac{1}{s^2}(-se^{-st}t-e^{-st}))_{0}^{\infty}$
$\Rightarrow{}\quad$ $\boxed{K(s)=\frac{4}{s^2}}$








(d) $m(t)=e^{2t}$

$\Rightarrow{}\quad$ $G(s)=\int_{0}^{\infty} \ {e^{-st}}{e^{2t}}{dt}={\left[-\frac{1}{s-2}{e^{-(s-2)t}}\right]_0^\infty}\quad$
$\Rightarrow{}\quad\ \boxed{G(s)=\frac{1}{s-2}}$








(e) $a(t)=\sin(5t)$

According to De Moivre's theorem, $\cos(at) + i\sin(at) = e^{iat}$
Using part (d) methodology $\Rightarrow{}\quad \mathcal{L}${$e^{5it}$} = $\frac{1}{s-5i}$
Rationalize the denominator $\Rightarrow{}\quad \mathcal{L}${$e^{5it}$} = $\frac{s}{s^2-5^2} + \frac{5i}{s^2-5^2}$
$\sin{}$ is the imaginary part. By using linearity $\Rightarrow{}\quad \boxed{A(s)=\frac{5}{s^2-5^2}}$
Note: The same process can be used to find the laplace of $\cos{}$ - take the real part, $\frac{s}{s^2+5^2}$!








Problem 2.

Using the DE1-MEM Formula sheet, find the Laplace Transforms of the given functions:

(a) $f(t)=6e^{-5t}+e^{3t}-5t^{3}-9$

$\Rightarrow{}\quad$ $F(s)=\mathcal{L}\{ {6e^{-5t}+e^{3t}-5t^{3}-9}\}$
$\Rightarrow{}\quad$ $\boxed{F(s)=\frac{6}{s+5}+\frac{1}{s-3}-\frac{30}{s^4}-\frac{9}{s}}$








(b) $g(t)=4\cos(4t)-9\sin(4t)+2\cos(10t)$

$\Rightarrow{}\quad$ $G(s)=\mathcal{L}\{ {4\cos(4t)}\}-\mathcal{L}\{ {9\sin(4t)}\}+\mathcal{L}\{ {2\cos(10t)}\}$
$\Rightarrow{}\quad$ $\boxed{G(s)=\frac{4s}{ {s^2}+{16}}-\frac{36}{ {s^2}+{16}}+\frac{2s}{ {s^2}+{100}}}$








(c) $k(t)=3\sinh(2t)+4\cosh(3t)$

$\Rightarrow{}\quad$ $K(s)=\mathcal{L}\{ {3\sinh(2t)}\}+\mathcal{L}\{ {4\cosh(3t)}\}$
$\Rightarrow{}\quad$ $\boxed{K(s)=\frac{6}{ {s^2}-{4}}+\frac{4s}{ {s^2}-{9}}}$








(d) $m(t)=e^{3t}+\cos(6t)-e^{3t}\cos(6t)$

$\Rightarrow{}\quad$ $M(s)=\mathcal{L}\{ {e^{3t}}\}+\mathcal{L}\{ {\cos(6t)}\}-\mathcal{L}\{ {e^{3t}\cos(6t)}\}$
$\Rightarrow{}\quad$ $\boxed{M(s)=\frac{1}{ {s}-{3}}+\frac{s}{ {s^2}+{36}}-\frac{s-3}{ {(s-3)^2}+{36}}}$








(e) $o(t)=e^{-2t}{\cos^2}(3t)-3t^2e^{3t}$

$\Rightarrow{}\quad$ ${\cos^2}(3t)$ can be writen as = $\frac{1+{\cos}(6t)}{2}$
$\Rightarrow{}\quad$ $O(s)=\mathcal{L}\{ {e^{-2t}}{(\frac{1}{2}+\frac{1}{2}{\cos}(6t))}-{3t^2e^{3t}}\}=\mathcal{L}\{\frac{1}{2}{e^{-2t}}+{\frac{1}{2}{e^{-2t}}{\cos}(6t)}-{3t^2e^{3t}}\}$
$\Rightarrow{}\quad$ $\boxed{O(s)=\frac{1}{ {2s}+{4}}+\frac{s+2}{2(s+2)^{2}+72}-\frac{6}{ {(s-3)^3}}}$








Problem 3.

Compute the inverse Laplace Transform of the given functions:

(a) $F(s)= \frac{5}{s}-\frac{4}{s-2}+\frac{24}{(s-2)^5}$

$\mathcal{L}^{-1}\{F(s)\}= \ \mathcal{L}^{-1}\{ {\frac{5}{s}-\frac{4}{s-2}+\frac{24}{(s-2)^5}}\}$
$\Rightarrow{}\quad$ $\boxed{f(t)=5-{4e^{2t}}+e^{2t}{t^4}}$








(b) $G(s)= \frac{7s-1}{(s+1)(s+2)(s-3)}$

Using partial fractions
$\mathcal{L}^{-1}\{G(s)\}= \ \mathcal{L}^{-1}\{ {\frac{7s-1}{(s+1)(s+2)(s-3)}}\}$
$\frac{7s-1}{(s+1)(s+2)(s-3)}= \frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{s-3}$
$\Rightarrow\quad\frac{2}{s+1}+\frac{-3}{s+2}+\frac{1}{s-3}$
Using WolframAlpha
Link to WolframAlpha $\Rightarrow{}\quad\ \boxed{g(t)=2e^{-t}-3e^{-2t}+e^{3t}}$








(c) $K(s)= \frac{4s+5}{(s-2)^2(s+3)}$

Using partial fractions
$\mathcal{L}^{-1}\{K(s)\} = \mathcal{L}^{-1}\{\frac{4s+5}{(s-2)^2(s+3)}\}$
$\frac{4s+5}{(s-2)^2(s+3)}=-\frac{7}{25(s+3)}+\frac{7}{25(s-2)}+\frac{13}{5(s-2)^2}$
Using WolframAlpha
Link to WolframAlpha $\Rightarrow{}\quad \ \boxed{k(t)=-\frac{7}{25}{e^{-3t}}+\frac{7}{25}{e^{2t}}+\frac{13}{5}{t}{e^{2t}}}$








Problem 4.

Solve the following ODE function using Laplace Transform:

$y”+4y’+8y=1\ \ if \ \ y(0)=0,\ \ y’(0)=0$

$\mathcal{L}\{y"(t)+4y'(t)+8y(t)\}=\mathcal{L}\{y"(t)\}+4\mathcal{L}\{y'(t)\}+8\mathcal{L}\{y(t)\}=\mathcal{L}\{1\}$
$\Rightarrow{}\quad \ s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+8Y(s)=\frac{1}{s}$
$\Rightarrow{}\quad \ (s^2+4s+8)Y(s)-(s+4)y(0)-y'(0)=\frac{1}{s}$
Substitute $y(0)=0,\ \ y'(0)=0$:
$\Rightarrow{}\quad \ (s^2+4s+8)Y(s)-0-0=\frac{1}{s}$
$\Rightarrow{}\quad \ (s^2+4s+8)Y(s)=\frac{1}{s}$
$$Y(s)=\frac{1}{s(s^2+4s+8)}$$ Find the inverse Laplace transform of $Y(s)$:
Using partial fractions
$\frac{1}{s(s^2+4s+8)}=\frac{A}{s}+\frac{Bs+C}{(s^2+4s+8)}$ $\quad\Rightarrow{}\ 1=A(s^2+4s+8)+(Bs+C)s$
$Y(s)=\frac{1}{8s}-{\frac{s+4}{8(s^2+4s+8)}}$
$Y(s)=\frac{1}{8s}-{\frac{1}{8} \frac{(s+2)}{(s+2)^2+4}-\frac{1}{8} \frac{2}{(s+2)^2+4}}$
Using WolframAlpha
In this case, WolframAlpha does not return a convenient rearrangement. Further working must be done.
Link to WolframAlpha $\frac{1}{8s}$ is in a form that can be transformed using the DE1 formula sheet. Further manipulation must be done on $\frac{-s-4}{8(s^2+4s+8)}$. By thinking ahead and shifting into a form that allows us to compare with the formula sheet, $\frac{-(s+2)-2}{8(s^2+4s+8)}$.
$\Rightarrow{}\quad\frac{-(s+2)}{8(s^2+4s+8)} + \frac{-2}{8(s^2+4s+8)}$
$\Rightarrow{}\quad\frac{-1}{8}\frac{s+2}{(s+2)^2+4}-\frac{1}{8}\frac{2}{(s+2)^2+4}$
$Y(s)=\frac{1}{8s}-{\frac{1}{8} \frac{(s+2)}{(s+2)^2+4}-\frac{1}{8} \frac{2}{(s+2)^2+4}}$
Note: Recognising manipulations like this will come with practice and familiarisation with the transforms.
$\boxed{y(t)=\frac{1}{8}-{\frac{1}{8}}{e^{-2t}{\cos2t}}-\frac{1}{8}{e^{-2t}{\sin2t}}}$








Exam Style Questions

Problem 5.

Solve the following ODE function using Laplace Transform:

$y”+4y’+4y=6e^{-2t}\ \ if \ \ y(0)=-2,\ \ y’(0)=8$

$\mathcal{L}\{y"(t)+4y'(t)+4y(t)\}=\mathcal{L}\{y"(t)\}+4\mathcal{L}\{y'(t)\}+4\mathcal{L}\{y(t)\}=\mathcal{L}\{6e^{-2t}\}$
$\Rightarrow{}\quad \ s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+4Y(s)=\frac{6}{s+2}$
$\Rightarrow{}\quad \ (s^2+4s+4)Y(s)-(s+4)y(0)-y'(0)=\frac{6}{s+2}$
Substitute $y(0)=-2,\ \ y'(0)=8$:
$\Rightarrow{}\quad(s^2+4s+4)Y(s)+2(s+4)-8=\frac{6}{s+2}$
$\Rightarrow{}\quad(s^2+4s+4)Y(s)=\frac{6}{s+2}-2s$
$\Rightarrow{}\quad \ (s+2)^2Y(s)=\frac{6}{s+2}-{2s}$ $$Y(s)=\frac{6}{(s+2)^3}-\frac{2s}{(s+2)^2}$$ Find the inverse Laplace transform of $Y(s)$:
Using partial fractions
$Y(s)=\frac{6}{(s+2)^3}-\frac{2(s+2-2)}{(s+2)^2}$
$Y(s)=\frac{6}{(s+2)^3}-\frac{2s+4}{(s+2)^2}+\frac{4}{(s+2)^2}$
$Y(s)=\frac{6}{(s+2)^3}-\frac{2}{(s+2)}+\frac{4}{(s+2)^2}$
Using WolframAlpha
Link to WolframAlpha $y(t)=3e^{-2t}t^2-2{e^{-2t}}+4e^{-2t}t$
$\boxed{y(t)=(3t^2+4t-2){e^{-2t}}}$








Problem 6

The displacement of water in a pipe is represented by $av”+bv’+cv = P(t)$ where $v(t)$ is the displaced volume, and $P(t)$ is the pressure applied to the system.

(a) A pressure of $e^{-t}$ is applied, and at $t = 0$, the displaced volume is -1 and the flow is 0. Given $a = 0, b=1$ and $c = -1$, find an explicit formula for the laplace transform of the volume written in it’s simplest form.

$v'(t)-v(t) = e^{-t}$
$sV(t) - v(0) -1(V(t)) = \frac{1}{s+1}$
$v'(0) = 0, v(0) = -1$
$V(t)(s-1) - v(0) = \frac{1}{s+1}$
$V(t)(s-1) - (-1) = \frac{1}{s+1}$
$V(t) = \frac{\frac{1}{s+1}-1}{s-1}$
Using WolframAlpha:
Link to WolframAlpha $\boxed{V(s)=\frac{-1}{2(s+1)}-\frac{1}{2(s-1)}}$








(b) What is the displaced volume when $t = 4$?

Transform back to the original dimension using the inverse laplace transform.
$\mathcal{L}^{-1}${$\frac{-1}{2(s+1)}-\frac{1}{2(s-1)}$}
$= \frac{-1}{2}(e^t+e^{-t})$
At $t = 4, \boxed{v=-27.3082}$








Challenging Questions

Problem 7.

This question demonstrates a way to solve PDEs (coming in 3 topics!) utilising laplace. The one dimensional heat equation for the temperature $T(x, t)$ satisfies $\frac{\partial ^{2}T}{\partial x^{2}} = \frac{1}{\sigma}\frac{\partial T}{\partial t}$ where $t$ is time, $x$ is a spatial dimension and $\sigma$ is a positive constant.

The temperature $T(x, t)$ is subject to the following conditions:
i. $\lim_{x \to \infty} [T(x,t)] = 0$
ii. $T(0,t) = 1$
iii. $T(x,0) = 0$

Use Laplace transforms to show that
\(\mathcal{L} [{T(x,t)}] = \overline{T}(x,s) = \frac{1}{s} exp[-\sqrt{\frac{s}{\sigma}} x].\)

Taking laplace transforms in t (x is unaffected)
$\Rightarrow\quad \mathcal{L} [\sigma\frac{\partial ^{2}T}{\partial x^{2}}] = \mathcal{L} [\frac{\partial T}{\partial t}] $
$\Rightarrow\quad \sigma\frac{\partial ^{2}}{\partial x^{2}}\overline{T} = s\overline{T} - T(x,0)$
According to initial conditions, the last term is equal to 0 therefore:
$\Rightarrow\quad \frac{\partial ^{2}\overline{T}}{\partial x^{2}} = \frac{s'}{\sigma}\overline{T}$
It is a second order ODE in $\overline{T}$ with exponential solutions (think of the ODE lesson!).
$\overline{T}(x,s) = A(s)e^{-\sqrt{\frac{s}{\sigma}}x} + B(s)e^{-\sqrt{\frac{s}{\sigma}}x} $
Apply $T \rightarrow 0$ as $x \rightarrow\infty$
$\Rightarrow \overline{T} \rightarrow 0$ as $x \rightarrow\infty$
$\therefore B(s) = 0$ and $\overline{T}(x,s) = A(s)e^{-\sqrt{\frac{s}{\sigma}}x}$

Apply $T(0,t) = 1 \Rightarrow \overline{T}(0,s) = \frac{1}{s}$
$\therefore \frac{1}{s} = A(s)$

$\boxed{\overline{T}(x,s) = \frac{1}{s}e^{-\sqrt{\frac{s}{\sigma}}x}}$








WolframAlpha

You can use the formula sheet to figure out the correct inverse laplace transformations or use WolframAlpha - try typing in inverse laplace 2/(s+4)^5. This won’t work out nicely with all numbers though!

Click here for WolframAlpha link

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Next week, Fourier Series!