Applied the Laplace Transform of the Derivative on the Tank Filling System

Applied the Laplace Transform of the Derivative on the Tank Filling System
This is a very important transform because derivatives appear in the ODEs we wish to
solve.




Example . Solve first ODF using the Laplace transform
y? - y = e3t; y(0) = 2:
Solution Application of the Laplace transform leads to
sY (s) - y(0) - Y (s) = 1/(s – 3) ; where L[e3t]= 1 / S-3
Therefore,
Y(s) (s-1) -2 = 1/(s – 3) Y(s)(S-1) = 2+ 1 /( S-3)

1=A(s-3) + B(s-1)
Let s=1: A=-1/2
Let s=3: B=1/2

Using the table to find the inverse Laplace transform, we obtain

Example. Solve second ODF y?? - 3y? + 2y = e3t ; y(0) = 1; y?(0) = 0:
Solution: Applying the Laplace transform,
[s2Y(s)-sY(0)-Y (0)]-3[sY(s)-Y(0)]+2Y(s) = 1/(s-3)
s2Y(s)-s-0-3[sY(s)-1]+2Y(s) = 1/(s-3)
s2Y(s)-s-3sY(s)+3+2Y(s) = 1/(s-3)
Y(s)[s2 -3s+2] -s +3 = 1/(s-3)

To find the inverse Laplace transform we will need first simplify the expression for Y (s) using the partial fraction decomposition.


To find A, B and C here is especially simple. For example, for A multiply both sides
by (s – 3) and plug s = 3 into the expressions to obtain A = 1/ 2. In a similar way B = -2 and C = 5 /2.

Therefore, using the linearity of the inverse Laplace transform, we will find.

Converting this equation to the Laplace domain:
; if ? i(s) = a/s
?(s?o(s) + ?o(0))+ ?o(s) = K a/s ; ?o(0) = 0
?o(s) (?s +1) = K a/s

Ka=A(?s +1) +Bs
Let s=0 ; A=Ka and if s=-1/? ; B = -ka?

Laplace transform of the integers




x(s) s(s2-1) = 3s2-1

Let s=0 ; A=1
Let s=1 ; B=2


Thus we see that functions are transformed from the t-domain into the s-domain by the Laplace transform.
Quiz:
5y +4y = 2 ; y(0)=1
y +y = 3 ; y(0)=2
y +2y = e3t ; y(0)=-5