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=60I(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=0y(0)=1y'(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}.

 

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