Goals

• State 1st Traslation Theorem.
  
  
• Provide some examples on how to apply the theorem.
  
  
• Apply the theorem to solve ODEs.

Motivation:

Following the definition of the Laplace transform, we have that: $$\mathcal{L}\{e^{5t}t\}=\int_0^\infty e^{-st}e^{5t}t\,dt=\int_0^\infty te^{-(s-5)t}\,dt$$ $$\lim_{a\to\infty} -\frac{te^{-(s-5)t}}{s-5}-\frac{e^{-(s-5)t}}{(s-5)^2}\bigg|_0^a=\frac{1}{(s-5)^2}.$$

1st Traslation Theorem

Applying the Theorem:

Example (Direct Calculation):

Solution:

1. The shifted function of $f$ is $t^3$ and its transform is $\frac{3!}{s^4}.$ Therefore the transform of $f$ is $\frac{3!}{(s-5)^4}.$
   
2. The shifted function of $g$ is $\cos(4t)$ and its transform is $\frac{s}{s^2+4^2}.$ Therefore the transform of $g$ is $\frac{s+2}{(s+2)^2+4^2}.$
   
3. The shifted function of $h$ is the unit step function and its transform is $\frac{e^s}{s}.$ Therefore the transform of $h$ is $e^{s+2}\frac{1}{s+2}.$
   

Example (Inverse Calculation):

$\mathcal{L}^{-1}\left\{\frac{2s+5}{(s-3)^2}\right\}$

Solution:

Let's find out the partial fractions: $\frac{2s+5}{(s-3)^2}=\frac{A}{(s-3)}+\frac{B}{(s-3)^2}=\frac{2}{s-3}+\frac{-1}{(s-3)^2}.$ Now we notice that those are shifted functions, therefore we look for the corresponding transform and we multiply it by an exponential factor: $$\mathcal{L}^{-1}\left\{\frac{2s+5}{(s-3)^2}\right\}=\mathcal{L}^{-1}\left\{\frac{2}{s-3}\right\}+\mathcal{L}^{-1}\left\{\frac{-1}{(s-3)^2}\right\}=2e^{3t}-te^{3t}.$$

Example (Inverse Calculation):

$\mathcal{L}^{-1}\left\{\frac{s+5}{s^2+4s+6}\right\}$

Solution:

First we check if the denominator has real roots: $s^2+4s+6=0\implies s=\frac{-4\pm\sqrt{16-4(1)(6)}}{2}=-2\pm i.$ Therefore the roots are complex and we need to complete the square: $$s^2+4s+6=(s+2)^2+4.$$ Therefore the fraction is $$\frac{s+5}{(s+2)^2+4}=\frac{(s+2)}{(s+2)^2+4}+\frac{3}{(s+2)^2+4}$$ We use the formulaes to find the inverse: $$\mathcal{L}^{-1}\left\{\frac{(s+2)}{(s+2)^2+4}\right\}=e^{-2t}\cos(2t)\quad \mathcal{L}^{-1}\left\{\frac{3}{(s+2)^2+4}\right\}=\frac{3}{2}e^{-2t}\sin(2t).$$

Example (Solving IVP)

First we apply Laplace transform in both sides of the ODE. $$s^2\mathcal{L}\{y\}-sy(0)-y'(0)-6s\mathcal{L}\{y\}+6y(0)+9\mathcal{L}\{y\}=\mathcal{L}\{t^2e^{3t}\}$$ $$\implies (s^2-6s+9)\mathcal{L}\{y\}=2s+5+\frac{2}{(s-3)^3}$$ We simplify and check if the quadratic polynomial is factorizable: $$\mathcal{L}\{y\}=\frac{2s+17}{(s-3)^3}+\frac{2}{(s-3)^3(s-3)^2}$$ We find the inverse Laplace transform.

Example (Solving IVP)

First we apply Laplace transform in both sides fo the ODE $$s^2\mathcal{L}\{y\}-sy(0)-y'(0)+4s\mathcal{L}\{y\}-4y(0)+6\mathcal{L}\{y\}=\mathcal{L}\{e^{-t}\}$$ $$\implies (s^2+4s+6)\mathcal{L}\{y\}=\frac{1}{s+1}$$ We use partial fractions: $$\mathcal{L}\{y\}=\frac{1}{(s+1)(s^2+4s+6)}=\frac{1/3}{s-1}+\frac{s-1/3}{s^2+4s+6}$$ We construct the function using the inverse Laplace transform.