### 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 $Y(s)$:

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

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

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

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