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MATH6037: Properties of the Laplace Transform — Selected Exercise Solutions

March 31, 2011 8:23 pm

Question 1

Find the Laplace transform of the following functions.

(i)

$6\cos 4t+t^3$

Solution

Using linearity:

$\mathcal{L}\{6\cos 4t+t^3\}=6\mathcal{L}\{\cos 4t\}+\mathcal{L}\{t^3\}$ .

Now using the tables:

$=6\frac{s}{s^2+4^2}+\frac{3!}{s^4}=\frac{6s}{s^2+16}+\frac{6}{s^4}$ .

(iii)

$t^3e^{-t}$ .

Solution

This needs the First Shift Theorem, which states:

$\mathcal{L}\{f(t)e^{at}\}=F(s-a)$ ,

where $F(s)=\mathcal{L}\{f(t)\}$ . Now looking at what we have to transform:

$e^{-t}t^3=e^{(-1)t}t^3$ ,

clearly what we need to do is find the Laplace transform of $t^3$ , $F(s)$ — and then replace $s$ by $s-(-1)=s+1$ . Now the Laplace transform of $t^3$ , by the tables, is $F(s)=3!/s^4$ . Hence

$\mathcal{L}\{t^3e^{-t}\}=\frac{6}{(s+1)^4}$ .

Question 2

Find the Laplace transforms of the functions that satisfy the following differential equations.

(i)

$4\frac{dI}{dt}+12I=60$ , $I(0)=0$ .

Solution

Take the Laplace transform of both sides:

$\mathcal{L}\left\{4\frac{dI}{dt}+12I\right\}=\mathcal{L}\{60\}$ .

Now use linearity;

$4\mathcal{L}\left\{\frac{dI}{dt}\right\}+12\mathcal{L}\{I\}=60\mathcal{L}\{1\}$ .

Now consulting the tables (for the first differentiation theorem and the laplace transform of 1):

$4s\mathcal{L}\{I\}-4I(0)+12\mathcal{L}\{I\}=\frac{60}{s}$ .

Now use the boundary condition — $I(0)=0$ ;

$4s\mathcal{L}\{I\}+12\mathcal{L}\{I\}=\frac{60}{s}$

Solving for $\mathcal{L}\{I\}$ ;

$\Rightarrow \mathcal{L}\{I\}(4s+12)=\frac{60}{s}$

$\Rightarrow \mathcal{L}\{I\}=\frac{60}{s(4s+12)}=\frac{15}{s(s+3)}$ .

(ii)

$y''+2y'+4y=0$ ; $y(0)=1$ , $y'(0)$ .

Solution

Taking the Laplace transform of both sides (using linearity and the fact that the laplace transform of $0$ is $0$ — also note that $Y(s)=\mathcal{L}\{y\}$ ):

$\mathcal{L}\{y''\}+2\mathcal{L}\{y'\}+4Y(s)=0$ .

Now applying the differentiation theorems:

$s^2Y(s)-sy(0)-y'(0)+2(sY(s)-y(0))+4Y(s)=0$ .

Applying the boundary conditions:

$s^2Y(s)-s(1)-0+2(sY(s)-1)+4Y(s)=0$ .

Now all that remains is to solve for $Y(s)$ :

$s^2Y(s)-s+2sY(s)-2+4Y(s)=0$

$\Rightarrow s^2Y(s)+2sY(s)+4Y(s)=s+2$

$\Rightarrow Y(s)(s^2+2s+4)=s+2$

$\Rightarrow Y(s)=\frac{s+2}{s^2+2s+4}$ .

Posted by J.P. McCarthy

Categories: MATH6037

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2 Responses to “MATH6037: Properties of the Laplace Transform — Selected Exercise Solutions”

Hi JP,
Any chance you could post the solutions to the Laplace transforms of “2 sin 3t cos t” , “3 cos 4t sin 2t” just so I/we have a couple of more examples of using the (A+B) formula in a product of sin and cos situation.
Thanks,
Graham

By Graham Kiely on May 19, 2011 at 7:47 pm
1. Roger.
  
  O.K., the main principle is that the Laplace transform is linear — that is it handles sums — we can split sums and take out constants. Therefore if we can write terms of the form $a\sin A\cos B$ , etc. as sums of sines and cosines we are away with it.
  
  The formulae that convert products like these into sums is in the log tables. Taking the example of, say, $3\cos 4t\sin 2t$ , we use the formula
  
  $2\cos A\sin A=\sin(A+B)+\sin(A-B)$
  
  We can write $3\cos 4t\sin 2t=\frac{3}{2}\times2\cos 4t\sin 2t$ . Hence
  
  $\frac{3}{2}\times2\cos 4t\sin 2t=\frac{3}{2}(\sin 6t+\sin 2t)$ .
  
  Now we can use the linear nature of the Laplace transform:
  
  $\mathcal{L}\{3 \cos 4t \sin 2t\}=\frac{3}{2}\mathcal{L}\{\sin 6t\}+\frac{3}{2}\mathcal{L}\{\sin 2t\}.$
  
  Now using the tables to find the Laplace transforms of terms of the form $\sin kt$ and $\cos kt$ :
  
  $\mathcal{L}\{3 \cos 4t \sin 2t\}=\frac{3}{2}\frac{6}{s^2+36}+\frac{3}{2}\frac{s}{s^2+36}.$
  
  By J.P. McCarthy on May 20, 2011 at 4:33 pm

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