Goals

• Introduce the inverse of a Laplace transform.
  
  
• Introduce the relationship between the Laplace transform and the derivative of a function.
  
  
• Solve initial value problems using Laplace transforms.
  
  

Most commun Laplace transforms:

The next list has the Laplace transforms for the most commun functions

• $\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}}$
• $\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a}$
• $\mathcal{L}\left\{\sin(kt)\right\}=\frac{k}{s^2+k^2}$
• $\mathcal{L}\left\{\cos(kt)\right\}=\frac{s}{s^2+k^2}$
• $\mathcal{L}\left\{\sinh(kt)\right\}=\frac{k}{s^2-k^2}$
• $\mathcal{L}\left\{\cos(kt)\right\}=\frac{s}{s^2-k^2}$

Inverse Laplace Transform

• $F(s)=\frac{1}{s^5}$
• $F(s)=\frac{1}{s^2+7}$
• $F(s)=\frac{s^2+6s+9}{(s-1)(s-2)(s+4)}$

Laplace transform of a Derivative:

Example:

Find the solution of the initial value problem $y'+3y=13\sin(2t)\quad y(0)=6.$

Solution:

1. We apply Laplace transform in both sides of the equation:

$$\mathcal{L}(y')+3\mathcal{L}(y)=\mathcal{13\sin(2t)}$$

1. We use derivative theorem to express the transform of the derivative.

$$s^2 \mathcal{L}(y)-sy(0)=\frac{26}{s^2+4}$$

1. We solve for $\mathcal{L}(y)$ $$(s^2-6s)\mathcal{L}(y)=\frac{26}{s^2+4}\Rightarrow \mathcal{L}(y)=\frac{26}{s(s-6)(s^2+4)}$$

Solution:

1. We use partial fractions to express $\mathcal{L}(y)$ as a sum of simpler fractions

$$\mathcal{L}(y)=\frac{13 (3 s - 2)}{40 (s^2 + 4)} + \frac{13}{120 (s - 6)} - \frac{13}{12 s}$$

1. Then we use the table of Laplace transforms to find the inverse transform.

$$y(t)=-\frac{13}{12}+\frac{13}{120}e^{6t}+\frac{13}{40}\left(3\cos(2t)-\frac{1}{2}\sin(2t)\right)$$

Example

Find the solution of the following initial value problem $$y''-3y'+2y=e^{-4t},\quad y(0)=1,\quad y'(0)=5.$$

Example

Find the solution of the following initial value problem $$2y'''+3y''-3y'-2y=e^{-t},\, y(0)=0,\,y'(0)=0,\,y''(0)=1.$$